Georgian Mathematical Journal Volume 8 (2001), Number 3, 571–614

BOUNDARY INTEGRAL EQUATIONS OF PLANE ELASTICITY IN DOMAINS WITH PEAKS

V. G. MAZ’YA AND A. A. SOLOVIEV

In memoriam N. I. Muskhelishvili

Abstract. Boundary integral equations of elasticity theory in a plane domain with a peak at the boundary are considered. Solvability and uniqueness theorems as well as results on the asymptotic behaviour of solutions near the peak are obtained.

2000 Mathematics Subject Classification: 31A10, 45A05.

Key words and phrases: Boundary integral equations, elastic potential.

1. Introduction

The theory of elastic potentials for domains with smooth boundaries is well developed (see the monographs [6], [17]). For domains with piecewise smooth boundaries “without zero angles” theorems on the unique solvability of integral equations of elasticity were obtained in [7] by a method which does not use Fredholm and singular integral operators theories. Solutions of integral equations are expressed by the inverse operators of auxiliary exterior and interior boundary value problems, i.e., theorems on the solvability of boundary integral equations follow from the theory of elliptic boundary value problems in domains with piecewise smooth boundaries.

We apply the same approach to integral equations of the plane elasticity theory on a contour with a peak. We also use the complex form of solutions to the elasticity equations suggested by G. V. Kolosov. This method was further developed by N. I. Muskhelishvili (see [16]).

Since even for smooth functions in the right-hand side these integral equations, in general, have no solutions in the class of summable functions, we study modified integral equations for which theorems on the unique solvability prove to be valid.

Using the same method we obtained (see [9]–[11]) asymptotic formulas for solutions of integral equations of the logarithmic potential theory near cusps on boundary curves. This approach permitted us also to find, for each integral equation, a pair of weighted L_{p}-spaces such that the corresponding integral operator maps one space onto another (see [12]–[15]).

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

In the recent articles [1], [2], criteria of solvability in weighted L_{p}-spaces of boundary integral equations of the logarithmic potential theory on contours with peaks were obtained. The method used in these papers is based on reducing of boundary value problems to the Riemann–Hilbert problem for analytic functions on the unit circumference.

Here we give a brief description of the results obtained in the present paper. Let Ω be a plane simply connected domain bounded by a closed piecewise

smooth curve S with a peak at the origin O. Suppose that either Ω or its complement Ω^{c} is described in the Cartesian coordinates x, y near O by the inequalities κ_{−}(x) < y < κ_{+}(x), 0 < x < δ, where κ_{±} are C^{∞}-functions on [0, δ] satisfying

κ_{±}(0) = κ′_{±}(0) = 0 and κ+^{′′}(0) > κ−^{′′}(0).

In the first case we say that O is an outward peak and in the second one O is an inward peak.

We introduce the class 𝔑_{ν} (ν > −1) of infinitely differentiable on S\{O} vector-valued functions h admitting representations h_{±}(x) = x^{ν}q_{±}(x) on the arcs S_{±} = {(x, κ_{±}(x)) : x ∈ (0, δ]}, where the vector-valued functions q_{±} belong to C^{∞}[0, δ] and satisfy |q_{+}(0)| + |q_{−}(0)| ≠ 0. Let 𝔑 denote the set

𝔑 = ⋃_{ν>−1} 𝔑_{ν},

and let 𝔐_{β} (β > −1) be the class of differentiable vector-valued functions on S\{O} satisfying

σ^{(r)}(z) = O(x^{β−r}), z = x + iy = (x, y), r = 0, 1.

We introduce the class 𝔐 as

𝔐 = ⋃_{β>−1} 𝔐_{β}.

For domains with an outward peak we put

𝔐_{ext} = ⋃_{β>−1/2} 𝔐_{β} .

We consider the interior and exterior first boundary value problems

△^{∗}u = μ△u + (λ + μ)∇div u = 0 in Ω, u = g on S ,  (D^{+})
△^{∗}u = 0 in Ω^{c}, u = g on S , u(z) = O(1) as |z| → ∞ ,  (D^{−})

and the interior and exterior second boundary value problems

△^{∗}u = 0 in Ω, T u = h on S ,  (N^{+})
△^{∗}u = 0 in Ω^{c}, T u = h on S , u(z) = o(1) as |z| → ∞ , (N^{−})

for the displacement u = (u_{1}, u_{2}). Here T (∂_{ζ}, n_{ζ}) is the traction operator

T (∂_{ζ}, n_{ζ}) u = 2μ∂u/∂n + λn div u + μ[n, rot u],

where n = (n_{ξ}, n_{η}) is the outward normal to the boundary S at the point ζ = (ξ, η), and λ, μ are the Lamé coefficients. Henceforth we shall not distinguish a displacement u = (u_{1}, u_{2}) and a complex displacement u = u_{1} + iu_{2}.

A classical method for solving the first and second boundary value problems of elasticity theory consists in representing their solutions in the form of the double-layer potential

W σ(z) = ∫_{S} {T (∂_{ζ}, n_{ζ}) Γ (z, ζ)}^{∗} σ(ζ)ds_{ζ} ,

and the simple-layer potential

V τ(z) = ∫_{S} Γ(z, ζ)τ(ζ)ds_{ζ} , z = (x, y) ∈ Ω or Ω^{c} ,

where * denotes the passage to the transposed matrix and Γ is the Kelvin– Somigliana tensor

Γ(z, ζ) =\frac{λ + 3μ}{4πμ(λ + 2μ)} \{log \frac{1}{|z − ζ|} (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix})

+ \frac{λ + μ}{λ + 3μ} \frac{1}{|z − ζ|^{2}} (\begin{matrix}(x − ξ)^{2} & (x − ξ)(y − η) \\ (x − ξ)(y − η) & (y − η)^{2} \end{matrix})\} .

For the problems D^{+} and N^{−} the densities of the corresponding potentials can be found from the systems of boundary integral equations

−2^{−1}σ + W σ = g  (1)

and

−2^{−1}τ + T V τ = h .  (2)

Under certain general conditions on g in (1) there exist solutions u^{+} and u^{−} of the problems D^{+} and D^{−} in Ω and Ω^{c} with the boundary data g satisfying

g(z) = lim_{ε→0} ∫_{{S:|ζ|>ε}} Γ(z, ζ) (T (∂_{ζ}, n_{ζ})u^{+}(ζ) − T (∂_{ζ}, n_{ζ})u^{−}(ζ))ds_{ζ} + u^{−}(∞) (3)

on S \ {O}. Let v^{−} denote a solution of N^{−} in Ω^{c}, vanishing at infinity, with the boundary data T u^{+} on S \ {O}. We can choose v^{−} so that, for w = v^{−} − u^{−} + u^{−}(∞) on z ∈ S \ {O}, the equality

w(z) − 2 lim_{ε→0} ∫_{{S:|ζ|>ε}} {T (∂_{ζ}, n_{ζ}) Γ (z, ζ)}^{∗}w(ζ)ds_{ζ} = −2φ(z) + 2u^{−}(∞) (4)

holds. Solutions of equations (1) and (2) are constructed by means of (3) and (4). So, the function

σ = v^{−} − g

is a solution of (1). A solution of (2) can be obtained as follows. Let us introduce the solution v^{−} of N^{−} in Ω^{c} with the boundary data h, vanishing at infinity, and the solution u^{+} of D^{+} in Ω equal to v^{−} on S \ {O}. Under sufficiently general assumptions on h we can select v^{−} and u^{+} so that the density

τ = T u^{+} − h

satisfies (2).

Inward peak. In fact, the integral equation (1), in general, has no solutions in 𝔐 even if g ∈ 𝔑 vanishes on S_{±}. However, for a function from 𝔑_{ν} with ν > 3 the solvability of (1) can be attained by changing the equation in the following way. A solution u of the problem D^{+} is sought as the sum of the double-layer potential with density σ and the linear combination of explicitly given functions A_{1}, A_{2} and A_{3} with unknown real coefficients

u(z) = W σ(z) + c_{1}A_{1}(z) + c_{2}A_{2}(z) + c_{3}A_{3}(z).

The functions A_{1}, A_{2}, A_{3} are given by

A_{1}(z) = \frac{i}{2μ}[2κ Im z^{1/2} − \overline{z^{−1/2}} Im z] +

+ i\frac{(κ − 1)(α_{+} − α_{−})}{8πκμ}[2κ Im (z^{3/2} log z) − 3\overline{z^{1/2} log z} Im z −

− 2\overline{z^{1/2}} Im z] − i\frac{(κ − 1)α_{+}}{2μ} \overline{z^{3/2}},

A_{2}(z) = −\frac{Q}{2μ}[2κ Im z^{1/2} + \overline{z^{−1/2}} Im z] −

− Q\frac{(α_{+} − α_{−})(κ + 1)}{8πμκ}[2κ Im (z^{3/2} log z) + 3\overline{z^{1/2} log z} Im z +

+ 2\overline{z^{1/2}} Im z] + \frac{i}{2μ}[2κ Im z^{3/2} − 3\overline{z^{1/2}} Im z] + Q\frac{(κ + 1)α_{+}}{2μ} \overline{z^{3/2}},

A_{3}(z) = −\frac{κ + 1}{μ} Im z − \frac{(α_{+} − α_{−})(κ + 1)}{4πμκ}[2κ Im (z^{2} log z) +

+ 4\overline{z log z} Im z + 2\overline{z} Im z] + \frac{(κ + 1)α_{+}}{μ} \overline{z^{2}},

where

κ = (λ + 3μ)/(λ + μ) and Q = [(α_{+} + α_{−}) − (α_{+} − α_{−})/2κ + 2α_{−}] /2 .

Here and in the sequel by symbols z^{ν}(log z)^{k} we mean the branch of the analytic function taking real values on the upper boundary of the slit along the positive part of the real axis. By the limit relation for the double-layer potential we obtain

−2^{−1}σ + W σ + c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3} = g  (5)

for the pair (σ, c), where c = (c_{1}, c_{2}, c_{3}).

We prove the uniqueness assertion for equation (5) in the class of pairs {σ, c} with σ ∈ 𝔐 and c ∈ R^{3} in Theorem 5. The solvability of (5) with the righthand side g ∈ 𝔑_{ν}, ν > 3, in 𝔐 × R^{3} is proved in Theorem 6. Moreover, in the same theorem we derive the following asymptotic formula for σ near the peak:

σ(z) = (α(log x)^{2} + β log x + γ) x^{−1/2} + O(x^{−ε}), z ∈ S,

with positive ε.

A solution v of the problem N^{−} with the boundary data h from 𝔑_{ν}, ν > 3, is sought in the form of the simple-layer potential V τ. The density τ satisfies the system of integral equations (2) on S\{O}. In Theorems 7 and 8 we prove that if h has the zero mean value on S, then equation (2) has the unique solution τ in the class 𝔐 and this solution admits the following representation on the arcs S_{±}:

τ_{±}(z) = α_{±}x^{−1/2} + O(1) .

Outward peak. We represent a solution u of the problem D^{+} as the doublelayer potential W σ. The density σ is found from the system of integral equations (1). It is proved that the kernel of the integral operator in (1) is two-dimensional in the class 𝔐. Solutions of the homogeneous system of integral equations (1) are functions obtained as restrictions to S of solutions to the homogeneous problem N^{−}. Near the peak these displacements have the estimate O(r^{−1/2}) with r being the distance to the peak. So 𝔐_{ext} is the uniqueness class for equation (1). The situation where (1) has at least two solutions is considered in Theorem 9.

The non-homogeneous integral equation (1) is studied in Theorem 10. We show that the solvability in 𝔐 holds for all functions g from the class 𝔑_{ν}, ν > 0. One of the solutions of (1) has the representations on S_{±}:

σ(x) = β_{±}x^{ν−1} + O(1) for ν ≠ 1/2,
σ(x) = β_{±}x^{−1/2} log x + O(1) for ν = 1/2.

The integral equation (2) is uniquely solvable in the class 𝔐 if the right-hand side h ∈ 𝔑_{ν} with ν > 0 satisfies

∫_{S} hds = 0, ∫_{S} hζds = 0,

where ζ is any solution of the homogeneous equation (1) in the class 𝔐. In order to remove the orthogonality condition we are looking for a solution v of the problem N^{−} with the boundary data h from 𝔑 as the sum of the simplelayer potential V τ and the linear combination of functions ϱ_{1}(z), ϱ_{2}(z) with unknown coefficients

v(z) = V τ(z) + t_{1}ϱ_{1}(z) + t_{2}ϱ_{2}(z).

The functions ϱ_{k}(z), k = 1, 2, are defined by complex stress functions (complex potentials) φ_{k}(z), ψ_{k}(z):

ϱ_{k}(z) = \frac{1}{2μ} [κφ_{k}(z) − z\overline{φ′_{k}(z)} − \overline{ψ_{k}(z)}] ,

φ_{1}(z) = (\frac{zz_{0}}{z − z_{0}} )^{1/2} , ψ_{1}(z) = −\frac{3}{2} (\frac{zz_{0}}{z − z_{0}} )^{1/2} ,

φ_{2}(z) = i (\frac{zz_{0}}{z − z_{0}} )^{1/2} , ψ_{2}(z) = \frac{i}{2} (\frac{zz_{0}}{z − z_{0}} )^{1/2} ,

where z_{0} is a fixed point in Ω. The boundary equation

−2^{−1}τ + T V τ + t_{1}T ϱ_{1} + t_{2}T ϱ_{2} = h  (6)

is considered with respect to the pair (τ, t), where τ is the density of the simplelayer potential and t = (t_{1}, t_{2}) is a vector in R^{2}. In Theorems 11 and 12 we prove the existence and uniqueness of the solution of (6), respectively. In Theorem 12 we also study the asymptotic behaviour of solutions. We prove that for h ∈ 𝔑_{ν} with 0 < ν < 1 the density τ has the following representations on the arcs S_{±}:

τ(x) = β_{±}x^{ν−1} + O(x^{−1/2}) for 0 < ν < 1/2,
τ(x) = γ_{±}x^{−1/2} log x + β_{±}x^{−1/2} + O(log x) for ν = 1/2,
τ(x) = γ_{±}x^{−1/2} + β_{±}x^{ν−1} + O(log x) for 1/2 < ν < 1.

Assertions on the asymptotics of solutions to problems D^{+} and N^{−} are collected in Theorems 1–4.

2. Boundary Value Problems of Elasticity

We represent densities of integral equations of elasticity theory by means of solutions of certain auxiliary interior and exterior boundary value problems. The auxiliary results concerning such problems are collected in this section.

2.1. Asymptotic behaviour of solutions to the problem D^{+}. We introduce some notation to be used in the proof of the following theorem and elsewhere.

Let β > 0. As in [5], by W_{2,β}^{l}(G) we denote the weighted Sobolev space with the norm

(∑_{k=0}^{ℓ} ∫_{G} |∇^{k}(e^{βt}f)|^{2}dtdu)^{1/2} ,

where ∇^{k} is the vector of all derivatives of order k. By W̊_{2,β}^{l}(G) we mean the completion of C_{0}^{∞} in the W_{2,β}^{l}(G)-norm and let W_{2,β}^{0} = L_{2,β}.

Theorem 1. Let Ω have an outward peak. Suppose that g is an infinitely differentiable function on the curve S\{O} and let g have the following representations on the arcs S_{±}:

g_{±}(z) = ∑_{k=0}^{n+1} Q_{±}^{(k+1)}(log x)x^{k+ν} + O(x^{n+2+ν−ε}), z = x + iy, ν > −1,

where Q_{±}^{(j)} are polynomials of degree j and ε is a small positive number.

Suppose the above representations can be differentiated n + 2 times. Then the problem D^{+} has a solution u of the form

u(z) = \frac{1}{2μ} [κφ_{n}(z) − z\overline{φ′_{n}(z)} − \overline{ψ_{n}(z)}] + u_{0}(z), z ∈ Ω,  (7)

where ∇^{k}u_{0}(z) = O(|z|^{n−2k}) for k = 0, …, n and

φ_{n}(z) = ∑_{k=0}^{n+1} P_{φ}^{(k+2)}(log z)z^{ν+k−1},

ψ_{n}(z) = ∑_{k=0}^{n+1} P_{ψ}^{(k+2)}(log z)z^{ν+k−1} .

Here P_{φ}^{(j)} and P_{ψ}^{(j)} are polynomials of degree j.

Proof. (i) We are looking for a displacement vector u_{n} such that the vectorvalued function g_{n} = g − u_{n} belong to C^{∞}(S\{O}) and (g_{n})_{±}(x) = x^{ν}(q_{n})_{±}(x) , where (q_{n})_{±} are infinitely differentiable on [0, δ] and satisfy ∇^{k}(q_{n})_{±}(x) = O(x^{n+1−k−ε}), k = 0, …, n + 2, on the arcs S_{±} with ε being a small positive number.

To this end, we use the method of complex stress functions (see [16], Ch. II). The displacement vector u is related to complex potentials φ and ψ as follows:

2μu(z) = κφ(z) − z\overline{φ′(z)} − \overline{ψ(z)},

where functions φ and ψ are to be defined by the boundary data of the problem D^{+}.

It suffices to consider a function g(z) coinciding with A_{±}x^{ν}(log x)^{m} on S_{±}. We shall seek the functions φ and ψ in the form

φ(z) = z^{ν−1} ∑_{k=0}^{m} β_{k}(log z)^{m−k} + ε_{0}z^{ν}(log z)^{m},

ψ(z) = z^{ν−1} ∑_{k=0}^{m} γ_{k}(log z)^{m−k} + δ_{0}z^{ν}(log z)^{m}

for ν ≠ 1. There exist β_{k}, γ_{k}, ε_{0} and δ_{0} such that κφ(z)−z\overline{φ′(z)}−\overline{ψ(z)} restricted to S_{±} is equal to 2μA_{±}x^{ν}(log x)^{m} plus terms of the form c_{±}x^{i}(log x)^{j}, admitting the estimate O(x^{ν}(log x)^{m−1}).

We substitute expansions of φ and ψ in powers of x along S_{±} into the equation

\frac{1}{2μ}(κφ(z) − z\overline{φ′(z)} − \overline{ψ(z)}) = g(z), z = x + iy ∈ S .

Comparing the coefficients in x^{ν}(log x)^{m} and x^{ν−1}(log x)^{m} we obtain the system

\{ \begin{matrix}i(κ′′_{+}(0) − κ′′_{−}(0))(ν − 1)(κβ_{0} + (ν − 3)\overline{β_{0}} + \overline{γ_{0}} = 4μ(A_{+} − A_{−}) \\ κβ_{0} − (ν − 1)\overline{β_{0}} − \overline{γ_{0}} = 0 \end{matrix}

with respect to β_{0} and γ_{0}. Let us choose ε_{0} arbitrarily. Then δ_{0} is defined by the equation

κε_{0} − ν\overline{ε_{0}} − \overline{δ_{0}} = μ(A_{+} + A_{−})

− \frac{i}{4}(κ′′_{+}(0) + κ′′_{−}(0))(ν − 1)(κβ_{0} + (ν − 3)\overline{β_{0}} + \overline{γ_{0}}) .

If β_{k} are given, then γ_{k} (k ≥ 1) are found from the chain of equations

κβ_{k} − (ν − 1)\overline{β_{k}} − \overline{γ_{k}} − (m − k + 1)\overline{β_{k−1}} = 0 .

In the case ν = 1 we seek the functions φ and ψ in the form

φ(z) = ∑_{k=0}^{m+1} β_{k}(log z)^{m+1−k} + ε_{0}z(log z)^{m},

ψ(z) = ∑_{k=0}^{m+1} γ_{k}(log z)^{m+1−k} + δ_{0}z(log z)^{m} .

The coefficients β_{0} and γ_{0} are found from the system

\{ \begin{matrix}κβ_{0} − \overline{γ_{0}} = 0 , \\ i(m + 1)(κ′′_{+}(0) − κ′′_{−}(0))(κβ_{0} − \overline{β_{0}} + \overline{γ_{0}}) = 2μ(A_{+} − A_{−}) . \end{matrix}

Further, we choose ε_{0} arbitrarily and find δ_{0} from the equation

κε_{0} − \overline{ε_{0}} − \overline{δ_{0}} = (m + 1)\overline{β_{0}} + μ(A_{+} + A_{−}) − \frac{i}{4}(κ′′_{+}(0) + κ′′_{−}(0))(κβ_{0} − \overline{β_{0}} + \overline{γ_{0}}) .

Given β_{k}, we find γ_{k} (k ≥ 1) from the chain of equations

κβ_{k} − \overline{γ_{k}} = (m + 1 − k)\overline{β_{k−1}} .

(ii) By u^{(1)} we denote a vector-valued function equal to g_{n} on S\{O} and satisfying the estimates

u^{(1)}(z) = O(|z|^{n+1+ν−ε}), ∇^{k}u^{(1)}(z) = O(|z|^{n+ν−k−ε}), k = 1, …, n + 2 .

Let the vector-valued function u^{(2)} be the unique solution of the boundary value problem

△^{∗}u^{(2)} = −△^{∗}u^{(1)} in Ω, u^{(2)} ∈ W̊_{2}^{1}(Ω).  (8)

After the change of the variable z = ζ^{−1} (ζ = ξ + iη), equation (8) with respect to U^{(2)}(ξ, η) = u^{(2)}(\frac{ξ}{|ζ|^{2}} , −\frac{η}{|ζ|^{2}} ) takes the form

L(∂_{ξ}, ∂_{η}) U^{(2)} = Δ^{∗} U^{(2)} + L(∂_{ξ}, ∂_{η}) U^{(2)} = F^{(1)} in Λ,

where a curvilinear semi-infinite strip Λ is the image of Ω, L(∂_{ξ}, ∂_{η}) is the second order differential operator with coefficients having the estimate O(1/ξ) as ξ → +∞, and ∇^{k}F^{(1)}(ζ) = O(|ζ|^{−n−ν−2−k+ε}), k = 0, …, n.

Let ρ be a function from the class C_{0}^{∞}(R) vanishing for ξ < 1 and equal to 1 for ξ > 2, and let ρ_{r}(ξ) = ρ(ξ/r). Clearly,

ξ^{n}L(∂_{ξ}, ∂_{η})ξ^{−n}Ũ^{(2)} = Δ^{∗}Ũ^{(2)} + R(∂_{ξ}, ∂_{η})Ũ^{(2)},

where R(∂_{ξ}, ∂_{η}) is the second order differential operator with coefficients admitting the estimate O(1/ξ) as ξ → +∞. Therefore the boundary value problem

Δ^{∗}Ũ^{(2)} + ρ_{r}Ũ^{(2)} = F^{(2)} in Λ, Ũ^{(2)} = 0 on ∂Λ,

where

F^{(2)}(ξ, η) = ξ^{n}F^{(1)}(ξ, η) and ∇^{k}F^{(2)}(ξ, η) = O(ξ^{−2−ν−k+ε}), k = 0, …, n ,

is uniquely solvable in W̊_{2}^{1}(Λ) for large r. From the local estimate

∥Ũ^{(2)}∥_{W_{2}^{n+2}(Λ∩{ℓ−1<ξ<ℓ+1})} ≤ const(∥χF^{(2)}∥_{W_{2}^{n}(Λ)} + ∥χŨ^{(2)}∥_{L_{2}(Λ)}) , (9)

where χ belongs to C_{0}^{∞}(ℓ − 2, ℓ + 2) and equals to one in (ℓ − 1, ℓ + 1), and from the Sobolev embedding theorem it follows that the vector-valued function Ũ^{(2)} and its derivatives up to order n are bounded as ξ → ∞. We set

U^{(3)}(ξ, η) = ξ^{−n}Ũ^{(2)}(ξ, η) and ∇^{k}U^{(3)}(ξ, η) = O(ξ^{−n}), k = 0, …, n.

Clearly, U^{(3)} belongs to the space W̊_{2}^{1}(Λ) and satisfies

L(∂_{ξ}, ∂_{η})U^{(3)} = F^{(1)}

for ξ > 2r. Using a partition of unity and the same local estimate we obtain that U^{(2)} − U^{(3)} ∈ W̊_{2}^{1} ∩ W_{2}^{n+2}(Λ_{2r}), where Λ_{2r} = Λ ∩ {ξ > 2r}.

Let D(∂_{ξ}, ∂_{η}) denote the differential operator △^{∗} continuously mapping W̊_{2,β}^{1}∩ W_{2,β}^{n+2}(Π) into W_{2,β}^{n}(Π), where Π= {(ξ, η) : −κ′′_{+}(x)/2<η < −κ′′_{−}(x)/2}. Eigenvalues of the operator pencil D(ik, ∂_{η}) are nonzero roots of the equation

α^{2}k^{2} = κ(sinh αk)^{2},

where α = (κ′′_{+}(0) − κ′′_{−}(0))/2 and κ = (λ + 3μ)/(λ + μ). Since the operator D(∂_{ξ}, ∂_{η}) is the “limit” operator for L(∂_{ξ}, ∂_{η}) and since the real axis has no eigenvalues of D(ik, ∂_{η}), there exists β > 0 such that

U^{(2)} − U^{(3)} ∈ W_{2,β}^{n+2} ∩ W̊_{2,β}^{1}(Λ)

(cf. [5], [8]). Now, since U^{(2)} = U^{(3)} + (U^{(2)} − U^{(3)}), it follows from the Sobolev embedding theorem that

∇^{k}U^{(2)}(ξ, η) = O(|ξ|^{−n}) for k = 0, …, n .

Therefore from (i) and (ii) we find that the function u = u_{n} + u^{(1)} + u^{(2)} is a solution of the problem D^{+} and has the required representation (7) with u_{0} = u^{(1)} + u^{(2)}.

Corollary 1.1. Let g have the following representations on the arcs S_{±}:

g_{±}(x) = ∑_{k=0}^{n+1} q_{±}^{(k)}x^{k+ν} + O(x^{n+2+ν}), ν > −1,

with real coefficients q_{±}^{(k)}. Then the functions φ_{n} and ψ_{n} in (7) have the form

φ_{n}(z) = β_{0}z^{−1} + (β_{1,0} + β_{1,1} log z) + ∑_{k=2}^{n+1} β_{k}z^{k−1},

ψ_{n}(z) = γ_{0}z^{−1} + (γ_{1,0} + γ_{1,1} log z) + ∑_{k=2}^{n+1} γ_{k}z^{k−1}

for ν = 0,

φ_{n}(z) = (β_{0,0} + β_{0,1} log z) + ∑_{k=1}^{n+1} β_{k}z^{k}, ψ_{n}(z) = (γ_{0,0} + γ_{0,1} log z) + ∑_{k=1}^{n+1} γ_{k}z^{k}

for ν = 1, and

φ_{n}(z) = ∑_{k=0}^{n+1} β_{k}z^{k+ν−1}, ψ_{n}(z) = ∑_{k=0}^{n+1} γ_{k}z^{k+ν−1}

otherwise.

Theorem 2. Let Ω have an inward peak. Suppose g is an infinitely differentiable function on the curve S\{O} and its restrictions to the arcs S_{±} have the representations

g_{±}(z) = ∑_{k=0}^{n+1} Q_{±}^{(k+1)}(log x)x^{k+ν} + O(x^{n+ν+2−ε}), ν > −1,

where Q_{±}^{(j)} are polynomials of degree j and ε is a small positive number. Suppose that these representations can be differentiated n + 2 times. Then the problem D^{+} has a solution of the form

u(z) = \frac{1}{2μ}[κ(φ_{n}(z) + φ_{∗}(z)) − z(\overline{φ′_{n}(z) + φ′_{∗}(z)})
− \overline{(ψ_{n}(z) + ψ_{∗}(z))}] + u_{0}(z) ,  (10)

where ∇^{ℓ}u_{0}(z) = O(|z|^{n+[ν]+1−ℓ−ε}), ℓ = 1, …, n. The complex potentials φ_{n}, ψ_{n}, φ_{∗} and ψ_{∗} are represented as follows:

φ_{n}(z) = ∑_{k=0}^{n} P_{φ}^{(k+2)}(log z)z^{k+ν}, ψ_{n}(z) = ∑_{k=0}^{n} P_{ψ}^{(k+2)}(log z)z^{k+ν},

φ_{∗}(z) = ∑_{m=1}^{p} R_{φ,m}(log z)z^{m/2}, ψ_{∗}(z) = ∑_{m=1}^{p} R_{ψ,m}(log z)z^{m/2}.

Here P_{φ}^{(j)}, P_{ψ}^{(j)} are polynomials of degree j, R_{φ,m}, R_{ψ,m} are polynomials of degree [(m − 1)/2], and p = 2(n + [ν] + 1).

Proof. We are looking for a displacement vector u_{n} such that the vector-valued function g_{n} = g − u_{n} on S\{O} belong to C^{∞}(S\{O}) and ∇^{k}(g_{n})_{±}(z) = O(x^{n+ν+3−k}) for k = 1, …, n+2. We use the method of complex stress functions. It suffices to take g(z) equal to A_{±}x^{ν}(log x)^{m} on S_{±}. As in Theorem 1, we introduce the potentials

φ(z) = β_{m}z^{ν}(log z)^{m} and ψ(z) = γ_{m}z^{ν}(log z)^{m}

for ν ≠ m/2, m ∈ Z such that κφ(z) − z\overline{φ′(z)} − \overline{ψ(z)} on S_{±} is the sum of 2μA_{±}x^{ν}(log x)^{m} and terms of the form c_{±}x^{i}(log x)^{j}, admitting the estimate O(x^{ν}(log x)^{m−1}). We substitute the expansions of φ and ψ in powers of x along S_{±} into the equation

\frac{1}{2μ}(κφ(z) − z\overline{φ′(z)} − \overline{ψ(z)}) = g(z), z = x + iy ∈ S .

The coefficients β_{m} and γ_{m} are found from the system

\{ \begin{matrix}κβ_{m} − ν\overline{β_{m}} − \overline{γ_{m}} = 2μA_{+} \\ e^{4iπν}κβ_{m} − ν\overline{β_{m}} − \overline{γ_{m}} = 2μe^{2iπν}A_{−} . \end{matrix}

If ν = m/2 we seek the functions φ and ψ in the form

φ(z) = (β_{m,1}(log z)^{m+1} + β_{m,0}(log z)^{m})z^{ν},
ψ(z) = (γ_{m,1}(log z)^{m+1} + γ_{m,0}(log z)^{m})z^{ν} .

In this case β_{m,1} and γ_{m,1} are found from the system

\{ \begin{matrix}κβ_{m,1} − ν\overline{β_{m,1}} − \overline{γ_{m,1}} = 0 , \\ κβ_{m,1} + ν\overline{β_{m,1}} + \overline{γ_{m,1}} = iμ\frac{A_{+} − (−1)^{m}A_{−}}{π(m + 1)} . \end{matrix}

Finally, we choose β_{m,0} arbitrarily. Then γ_{m,0} is defined by the equation

κβ_{m,0} − ν\overline{β_{m,0}} − \overline{γ_{m,0}} = 2μA_{+} + (m + 1)\overline{β_{m,1}} .

(ii) Let u^{(1)} be a vector-valued function equal to g_{n} on S\{O} and admitting the estimates

u^{(1)}(z) = O(|z|^{n+ν+3}) and ∇^{k}u^{(1)}(z) = O(|z|^{n+ν+2−k}), k = 1, …, n + 2,

in a neighborhood of the peak. By u^{(2)} = (u_{1}^{(2)}, u_{2}^{(2)}) we denote the solution of the Dirichlet problem

△^{∗}u^{(2)} = −△^{∗}u^{(1)} in Ω, u^{(2)} ∈ W̊_{2}^{1}(Ω) .

Let Λ be the image of Ω under the mapping (r, θ) → (t, θ), where r, θ are polar coordinates of (x, y) and t = log(1/r). The vector-valued function U^{(2)}(t, θ) with the components

u_{1}^{(2)}(e^{−t}, θ) cos θ + u_{2}^{(2)}(e^{−t}, θ) sin θ and u_{2}^{(2)}(e^{−t}, θ) cos θ − u_{1}^{(2)}(e^{−t}, θ) sin θ,

is a solution of the equation

△^{∗}U^{(2)} + KU^{(2)} = F^{(1)} in W̊_{2}^{1}(Λ) ,

where F^{(1)}(t, θ) = O(e^{−(n+ν+2)t}). Here K is the first order differential operator

K = (\begin{matrix}−λ + 2μ & −(λ + 3μ)(∂/∂θ) \\ (λ + 3μ)(∂/∂θ) & −μ \end{matrix}) .

From the local estimate

∥U^{(2)}∥_{W_{2}^{n+2}(Λ∩{ℓ−1<ξ<ℓ+1})} ≤ const(∥χF^{(1)}∥_{W_{2}^{n}(Λ)} + ∥χU^{(2)}∥_{L_{2}(Λ)}) , (11)

where χ belongs to C_{0}^{∞}(ℓ − 2, ℓ + 2) and equals to 1 in (ℓ − 1, ℓ + 1), it follows that U^{(2)} ∈ W_{2}^{n+2} ∩ W̊_{2}^{1}(Λ).

By D(∂_{t}, ∂_{θ}) we denote the operator △^{∗} + K continuously mapping W̊_{2,β}^{1} ∩ W_{2,β}^{n+2}(Π) into W_{2,β}^{n}(Π), where Π = {(t, θ) : 0 < θ < 2π, t ∈ R}. Eigenvalues of the operator pencil D(ik, ∂_{θ}) are the numbers k = iℓ/2, where ℓ ∈ Z, ℓ ≠ 0. The multiplicity of each eigenvalue is equal to 2 and the maximum length of the Jordan chain for each eigenvector (multiplicity of eigenvector) is equal to 1. Therefore, the strip 0 < Im z < β, where β ∈ (n + [ν] + 1, n + [ν] + 3/2), contains p = 2(n + [ν] + 1) eigenvalues of D(ik, ∂_{θ}).

Since F^{(1)} ∈ W_{2,β}^{n}(Λ), U^{(2)} admits the representation (cf. [5], [8])

U^{(2)} = ∑_{k=1}^{p} c_{k}V^{(k)} + W^{(1)},

where V^{(k)} = (V_{1}^{(k)}, V_{2}^{(k)}) are linear independent vector-valued functions satisfying (△^{∗} + K)V^{(k)} = 0 in Λ_{R} = Λ ∩ {t > R} and vanishing on ∂Λ ∩ {t > R}, V^{(k)} ∉ W_{2,β}^{n+2}(Λ_{R}) and W^{(1)} ∈ W_{2,β}^{n+2}(Λ_{R}). Making the inverse change t = − log r we obtain

u^{(2)}(r, θ) = ∑_{k=1}^{p} c_{k}v^{(k)}(r, θ) + w^{(1)}(r, θ) ,

where v^{(k)}(r, θ) = V^{(k)}(log(1/r), θ) · e^{iθ} and ∇^{ℓ}w^{(1)}(r, θ) = O(r^{n+[ν]+1−ℓ}) for ℓ = 1, …, n. Using the method of complex stress functions and repeating the above-mentioned arguments we find

∑_{k=1}^{p} c_{k}v^{(k)}(z) = \frac{1}{2μ} [κφ_{∗}(z) − z\overline{φ′_{∗}(z)} − \overline{ψ_{∗}(z)}] + w^{(2)}(z) ,

where ∇^{ℓ}w^{(2)}(z) = O(|z|^{n+[ν]+1−ℓ−ε}) for ℓ = 1, …, n and

φ_{∗}(z) = ε_{1}z^{1/2} + ε_{2}z + (ε_{3,0} + ε_{3,1} log z)z^{3/2} + ⋯ + R_{φ, p−1}(log z)z^{n+[ν]+1/2},

ψ_{∗}(z) = δ_{1}z^{1/2} + δ_{2}z + (δ_{3,0} + δ_{3,1} log z)z^{3/2} + ⋯ + R_{ψ, p−1}(log z)z^{n+[ν]+1/2}.

It follows from (i) and (ii) that the vector-valued function u = u_{n} + u^{(1)} + u^{(2)} is the required solution of the problem D^{+} with u_{0} = u^{(1)} + w^{(1)} + w^{(2)}.

Corollary 2.1. Let g have the following representations on the arcs S_{±}:

g_{±}(z) = ∑_{k=0}^{n+1}(α_{±}^{(k,1)}log x + α_{±}^{(k,0)})x^{k+ν} + O(x^{n+ν+2})

for ν ≠ m/2, m ∈ Z, where α_{±}^{(k,i)} are real numbers. Then the functions φ_{n} and ψ_{n} in (10) have the form

φ_{n}(z) = ∑_{k=0}^{n}(β^{(k,1)}log z + β^{(k,0)})z^{k+ν},

ψ_{n}(z) = ∑_{k=0}^{n}(γ^{(k,1)}log z + γ^{(k,0)})z^{k+ν}

with β^{(k,i)}, γ^{(k,i)} ∈ C.

2.2. Asymptotic behaviour of solutions to the problem N^{−}. We introduce the weighted space W^{k,ρ}(Ω^{c}) with the inner product

(f_{1}, f_{2})_{k,ρ} := ∑_{|α|≤k} ∫_{Ω^{c}} ρ^{−2k+2|α|}D^{α}f_{1}\overline{D^{α}f_{2}}dxdy,

where ρ(z) = (1 + |z|^{2}). By W̊^{k,ρ}(Ω^{c}) we denote the completion of C_{0}^{∞}(Ω^{c}) in W^{k,ρ}(Ω^{c}).

Theorem 3. Let Ω have an inward peak. Suppose that h is an infinitely differentiable vector-valued function on S\{O}, ∫_{S} h ds = 0 and let the restriction of h to S_{±} admit the representation

h_{±}(z) = ∑_{k=0}^{n−1} H_{±}^{(k+1)}(log x)x^{k+ν} + O(x^{n+ν−ε}), ν > −1,

where H_{±}^{(j)} are polynomials of degree j and ε is a small positive number. Let this representation be differentiable n times. Then the problem N^{−} with the boundary data h has a solution v bounded at infinity, satisfying the condition

V.P. ∫_{S} T v ds = lim_{ε→0}  ∫_{{q∈S, |q|≥ε}} T v ds = 0 ,  (12)

and, up to a linear function α + icz with real coefficient c, represented in the form

v(z) = \frac{1}{2μ} [κφ_{n}(z) − z\overline{φ′_{n}(z)} − \overline{ψ_{n}(z)}] + v_{0}(z) .  (13)

Here ∇^{k}v_{0}(z) = O(|z|^{n−2k−1}) for k = 1, …, n − 1,

φ_{n}(z) = i( ∑_{m=0}^{p} β_{0,m} (log \frac{zz_{0}}{z_{0} − z} )^{m}) (\frac{zz_{0}}{z_{0} − z} )^{ν−2}

+ ∑_{k=1}^{n+1} P_{φ}^{(k+2)} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k+ν−2} ,

ψ_{n}(z) = i( ∑_{m=0}^{p} γ_{0,m} (log \frac{zz_{0}}{z_{0} − z} )^{m}) (\frac{zz_{0}}{z_{0} − z} )^{ν−2}

+ ∑_{k=1}^{n+1} P_{ψ}^{(k+2)} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k+ν−2} ,

where β_{0,m} and γ_{0,m} are real numbers, p = 1 if ν ≠ 0, 1, 2, 3, and p = 2 otherwise, P_{φ}^{(j)} and P_{ψ}^{(j)} are polynomials of degree j.

Proof. (i) We are looking for a displacement vector v_{n} such that the traction h_{n} = h − T v_{n} belong to C^{∞}(S\{0}) and admit the estimates

∇^{k}(h_{n})_{±}(z) = O(x^{n+ν−k−ε}), z = x + iy,

on S_{±} for k = 0, …, n. To this end we represent the boundary condition of the problem N^{−} with the boundary data h in the Muskhelishvili form (see [16], Ch. II, Sect. 30)

φ(z) + z\overline{φ′(z)} + \overline{ψ(z)} = f(z), z ∈ S\\{0\} .  (14)

Here φ and ψ are complex stress functions and f has the form

f(z) = −i ∫_{(0z)^{⌣}} h ds + const, z ∈ S ,

where by (0z)^{⌣} we denote the arc of S connecting 0 and z. As f in (14), it suffices to consider the function ±ih_{±}x^{ν+1}(log x)^{m} on S_{±}. In a small neighborhood of the peak, we are looking for complex potentials φ and ψ in the form

φ(z) = ∑_{r=0}^{3}(∑_{k=0}^{p} \frac{m!}{(m − k)!}β ′_{r,k}(log z)^{p−k})z^{ν+r−2}

and

ψ(z) = ∑_{r=0}^{3}(∑_{k=0}^{p} \frac{m!}{(m − k)!}γ ′_{r,k}(log z)^{p−k})z^{ν+r−2} ,

where p = m if ν ≠ 0, 1, 2, 3 and p = m + 1 otherwise. As in Theorem 1, we find coefficients of φ and ψ so that the restriction of φ(z) + z\overline{φ′(z)} + \overline{ψ(z)} to S_{±} is the sum of ±ih_{±}x^{ν+1}(log x)^{m} and terms of the form c_{±}x^{i}(log x)^{j}, admitting the estimate O(x^{ν+1}(log x)^{m−1}).

The coefficients β ′_{0,k}, γ ′_{0,k}, k = 0, …, m, are defined by the system

\{ \begin{matrix}β ′_{0,k} + (ν − 2)\overline{β ′_{0,k}} + \overline{γ ′_{0,k}} = A_{0,k} , \\ β ′_{0,k} − (ν − 4)\overline{β ′_{0,k}} − \overline{γ ′_{0,k}} = B_{0,k} , \end{matrix}

where A_{0,k} = B_{0,k} = 0 for k = 0 and A_{0,k} = −\overline{β ′_{0,k−1}}, B_{0,k} = \overline{β ′_{0,k−1}} if k = 1, …, m. We find

Re β ′_{0,k} = Re γ ′_{0,k} = 0, k = 0, …, m,

and

(3 − ν) Im β ′_{0,0} = Im γ ′_{0,0} ,
(3 − ν) Im β ′_{0,k} = Im γ ′_{0,k} + Im β ′_{0,k−1}, k = 1, …, m .

The coefficients β ′_{1,0} and γ ′_{1,0} satisfy

\{ \begin{matrix}β ′_{1,0} + (ν − 1)\overline{β ′_{1,0}} + \overline{γ ′_{1,0}} = 0 , \\ β ′_{1,0} − (ν − 3)\overline{β ′_{1,0}} − \overline{γ ′_{1,0}} = \frac{(ν − 3)(ν − 2)}{ν − 1}(κ′′_{+}(0) + κ′′_{−}(0)) Im β ′_{0,0} . \end{matrix}

Hence it follows that

Re β ′_{1,0} = \frac{(ν − 3)(ν − 2)}{4(ν − 1)}(κ′′_{+}(0) + κ′′_{−}(0)) Im β ′_{0,0},

Re γ ′_{1,0} = −ν Re β ′_{1,0} and Im γ ′_{1,0} = −(ν − 2) Im β ′_{1,0}.

We set

Im γ ′_{1,0} = Im β ′_{1,0} = 0.

The coefficients β ′_{1,1} and γ ′_{1,1} are found from the system

\{ \begin{matrix}β ′_{1,1} + (ν − 1)\overline{β ′_{1,1}} + \overline{γ ′_{1,1}} & = −\overline{β ′_{1,0}} , \\ β ′_{1,1} − (ν − 3)\overline{β ′_{1,1}} − \overline{γ ′_{1,1}} & = \overline{β ′_{1,0}} \\ & + \frac{(ν − 3)(ν − 2)}{ν − 1}(κ′′_{+}(0) + κ′′_{−}(0)) Im β ′_{0,1} \\ & + \frac{ν^{2} − 2ν − 1}{(ν − 1)^{2}}(κ′′_{+}(0) + κ′′_{−}(0)) Im β ′_{0,0} . \end{matrix} (15)

586 V. G. MAZ’YA AND A. A. SOLOVIEV

Summing up these equations we obtain

Re β ′_{1,1} = \frac{(ν − 3)(ν − 2)}{4(ν − 1)}(κ′′_{+}(0) + κ′′_{−}(0)) Im β ′_{0,1}
+\frac{ν^{2} − 2ν − 1}{4(ν − 1)^{2}}(κ′′_{+}(0) + κ′′_{−}(0)) Im β ′_{0,0},
Re γ ′_{1,1} = −ν Re β ′_{1,1} − Re β ′_{1,0}.

From the first equation of system (15) we find

Im γ ′_{1,1} = −(ν − 2) Im β ′_{1,1} − Im β ′_{1,0} .

We set

Im β ′_{1,1} = Im γ ′_{1,1} = 0 .

The real parts of β ′_{1,k} and γ ′_{1,k}, k = 2, …, m, are found from the systems

\{ \begin{matrix}β ′_{1,k} + (ν − 1)\overline{β ′_{1,k}} + \overline{γ ′_{1,k}} = A_{1,k} , \\ β ′_{1,k} − (ν − 3)\overline{β ′_{1,k}} − \overline{γ ′_{1,k}} = B_{1,k} , \end{matrix}

where

A_{1,k} = −\overline{β ′_{1,k−1}},
B_{1,k} = −\frac{1}{ν − 1} (β ′_{1,k−1} − 2(ν − 2)\overline{β ′_{1,k−1}} −\overline{γ ′_{1,k−1}} − β ′_{1,k−2})
+\frac{(κ′′_{+}(0) + κ′′_{−}(0))}{ν − 1} (Im β ′_{0,k−2} +(2ν − 5) Im β ′_{0,k−1}
+(ν − 3)(ν − 2) Im β ′_{0,k}) .

Let Im β ′_{1,k} = 0, k = 2, …, m. The coefficients Im β ′_{2,0} and Im γ ′_{2,0} are calculated by the system

\{ \begin{matrix}β ′_{2,0} + ν\overline{β ′_{2,0}} + \overline{γ ′_{2,0}} = A_{2,0} , \\ β ′_{2,0} − (ν − 2)\overline{β ′_{2,0}} − \overline{γ ′_{2,0}} = B_{2,0} , \end{matrix}  (16)

where

A_{2,0} = −\frac{1}{2}iκ′′_{+}(0)κ′′_{−}(0)(ν − 3)(ν − 2) Im β ′_{0,0},
B_{2,0} = 2\frac{h_{+} + h_{−}}{ν(κ′′_{+}(0) − κ′′_{−}(0))} + \frac{(ν − 2)(ν − 3)}{3ν}(κ′′′_{+}(0) + κ′′′_{−}(0)) Im β ′_{0,0}

− i\frac{(ν − 2)(ν − 3)(ν − 4)}{6ν}((κ′′_{+}(0))^{2} + κ′′_{+}(0)κ′′_{−}(0) + (κ′′_{−}(0))^{2}) Im β ′_{0,0}

+ \frac{(ν − 1)(ν − 2)}{ν}(κ′′_{+}(0) + κ′′_{−}(0)) Re β ′_{1,0} .

System (16) is solvable if

\frac{(ν − 2)(ν − 3)(ν + 2)}{3}(κ′′_{+}(0) − κ′′_{−}(0))^{3} Im β ′_{0,0} = −8 Im(h_{+} + h_{−}). (17)

We express Im β ′_{0,0} from (17). Summing up the equations of system (16) we find Re β ′_{2,0}. Then Re γ ′_{2,0} is defined from the first equation in (16)

Re γ ′_{2,0} = −(ν + 1) Re β ′_{2,0}.

We choose Im β ′_{2,0} arbitrarily and find Im γ ′_{2,0} from the equation

(ν − 1) Im β ′_{2,0} + Im γ ′_{2,0} = \frac{1}{2}κ′′_{+}(0)κ′′_{−}(0)(ν − 3)(ν − 2) Im β ′_{0,0} .

The coefficients β ′_{2,1} and γ ′_{2,1} are found by the equation

β ′_{2,1} + ν\overline{β ′_{2,1}} + \overline{γ ′_{2,1}} = −\frac{1}{2}iκ′′_{+}(0)κ′′_{−}(0)((2ν − 5) Im β ′_{0,0} + (ν − 3)(ν − 2) Im β ′_{0,1}) .

We set

Im β ′_{0,1} = Im γ ′_{0,1} = 0 and Re β ′_{2,1} = Re γ ′_{2,1} = 0.

Let β ′_{2,1} be chosen. Then γ ′_{2,1} is subject to

−(ν − 1) Im β ′_{2,1} − Im γ ′_{2,1} = \frac{1}{2}κ′′_{+}(0)κ′′_{−}(0)(2ν − 5) Im β ′_{0,0} .

Finally, given β ′_{2,k} arbitrarily, we define γ ′_{2,k}, k ≥ 2, recursively

β ′_{2,k} + ν\overline{β ′_{2,k}} + \overline{γ ′_{2,k}} = \frac{1}{2}iκ′′_{+}(0)κ′′_{−}(0) (Im β ′_{0,k−2} +(2ν − 5) Im β ′_{0,k−1}
+ (ν − 3)(ν − 2) Im β ′_{0,k}) .

We set

Im β ′_{0,k} = Im γ ′_{0,k} = 0 and Re β ′_{2,k} = Re γ ′_{2,k} = 0, k ≥ 2 ,

and choose Im β ′_{2,k} and Im γ ′_{2,k} to satisfy

−(ν − 1) Im β ′_{2,k} − Im γ ′_{2,k} = \frac{1}{2}κ′′_{+}(0)κ′′_{−}(0)(Im β ′_{0,k−2} + (2ν − 5) Im β ′_{0,k−1}) .

The exceptional cases ν = 0, 1, 2, 3 can be treated similarly. We set

φ_{n}(z) = ∑_{r=0}^{3}(∑_{k=0}^{p} β_{r,k} (log \frac{zz_{0}}{z_{0} − z} )^{p−k}) (\frac{zz_{0}}{z_{0} − z} )^{r+ν−2} ,

ψ_{n}(z) = ∑_{r=0}^{3}(∑_{k=0}^{p} γ_{r,k} (log \frac{zz_{0}}{z_{0} − z} )^{p−k}) (\frac{zz_{0}}{z_{0} − z} )^{r+ν−2} , z ∈ Ω^{c} ,

where z_{0} is a fixed point of Ω. We choose β_{r,k} and γ_{r,k} so that in the decompositions of φ_{n} and ψ_{n} along S_{±} the coefficients in x^{ν−k}(log x)^{m}, k = 0, 1, 2, m = 1, …, p, coincide with the corresponding coefficients in the decompositions of φ and ψ.

The displacement vector v_{n} is defined by

v_{n}(z) = (2μ)^{−1} (κφ_{n}(z) − z\overline{φ ′_{n}(z)} − \overline{ψ_{n}(z)}) .

Then ∇v_{n} = O(|z|^{−2}) as |z| → ∞ and satisfies V.P. ∫_{S} T v_{n} ds = 0.

(ii) Let f_{n,1} and f_{n,2} denote the components of the vector-valued function f_{n} on S\{O} defined by

f_{n}(z) = −i ∫_{(0z)^{⌣}} h_{n} ds + const.

We set

A_{n} = −\frac{i}{2π} ∫_{S} f_{n,1}dx + f_{n,2}dy

and consider the function

χ_{n}(z) = A_{n} (log (1 − \frac{z}{z_{0}} ) + ∑_{k=1}^{n+[ν]+1} c_{k} (\frac{zz_{0}}{z_{0} − z} )^{k}) , z ∈ Ω^{c},

where z_{0} is a fixed point in Ω and the coefficients c_{k} are chosen so that χ_{n}(z) = O(|z|^{n+[ν]+2}) as z tends to zero. If by v_{n}^{(1)} we denote the displacement vector v_{n}^{(1)}(z) = (1/2μ)\overline{χ′(z)}, then T v_{n}^{(1)} = i(∂/∂s)\overline{χ′_{n}}. For the traction h_{n}^{(1)} = h_{n}−T v_{n}^{(1)} on S\{O} with the components h_{n1}^{(1)} and h_{n2}^{(1)} the principal vector ∫_{S} h_{n}^{(1)}(z)ds and the principal moment ∫_{S} [(x−x_{0})h_{n2}^{(1)}(x, y)−(y−y_{0})h_{n1}^{(1)}(x, y)]ds with respect to z_{0} = (x_{0}, y_{0}) ∈ Ω are equal to zero.

We need to construct a displacement vector v_{n}^{(2)} with the given stress h_{n}^{(1)} on S. To this end, we represent the boundary conditions of this problem via the Airy function F (z) biharmonic in Ω^{c} (see [16], Ch. II, Sect. 30). We have F = b and ∂F/∂n = d on S\{O}, where the functions b and d belong to C^{∞}(S\{O}) and admit the estimates

b(p) = O(|p|^{n+2+ν−ε}), d(p) = O(|p|^{n+1+ν−ε}), p ∈ S\{O} ,

because of their relation with h_{n}^{(1)}:

F (z) = ∫_{(0z)^{⌣}} [(x_{s} − x)h_{n1}^{(1)}(x_{s}, y_{s}) − (y_{s} − y)h_{n2}^{(1)}(x_{s}, y_{s})]ds, z = (x, y) ∈ S\{O},

\frac{∂F}{∂x} (z) = − ∫_{(0z)^{⌣}} h_{n2}^{(1)}ds + C_{1}, \frac{∂F}{∂y} (z) = ∫_{(0z)^{⌣}} h_{n1}^{(1)}ds + C_{2}.

(16)

Let F_{1}(z) in Ω^{c} satisfy

F_{1}(p) = b(p), (∂F_{1}/∂n)(p) = d(p) on S\{O},

and admit the estimate

∇^{k}F_{1}(z) = O(|z|^{n+1+ν−k−ε}), k = 0, …, n + 2, as z → 0 .

We can assume also that ρ^{−2}Δ^{2}F_{1} ∈ L_{2}(Ω^{c}). Then the boundary value problem

△^{2}F_{2} = −Δ^{2}F_{1} in Ω^{c}, F_{2} = ∂F_{2}/∂n = 0 on S,

is uniquely solvable in W̊^{2,ρ}(Ω^{c}) (cf. [4]).

We use the Kelvin transform setting

U(ζ) = |ζ|^{2}F_{2}(1/ζ), ζ = ξ + iη ∈ Λ,

where Λ is the image of Ω^{c} under the inversion ζ = 1/z. We have △^{2}U = H in Λ, where H(ζ) = O(|ζ|^{−n−ν−3+ε}). As in Theorem 1, we can construct a vectorvalued function U_{n}^{(1)} ∈ W̊_{2}^{2}(Λ ∩ {ξ > 2R}) such that ∇^{k}U_{n}^{(1)}(ξ, η) = O(ξ^{−n+1}), k = 0, …, n, △^{2}U_{n}^{(1)} = H in Λ ∩ {ξ > 2R} for large R and U − U_{n}^{(1)} belongs to (W̊_{2}^{2} ∩ W_{2}^{n+2})(Λ ∩ {ξ > 2R}).

By U(∂_{ξ}, ∂_{η}) we denote the operator △^{2} continuously mapping W̊_{2,β}^{2}∩W_{2,β}^{n+2}(Π) into W_{2,β}^{n−2}(Π), where Π = {(ξ, η) : −κ′′_{+}(x)/2 < η < −κ′′_{−}(x)/2}. The eigenvalues of the operator pencil U(ik, ∂_{η}) are nonzero roots of the equation

(αk)^{2} − (sinhαk)^{2} = 0,

where α = (κ′′_{+}(0) − κ′′_{−}(0))/2. Since the real axis contains no eigenvalues of U(ik, ∂_{η}), there exists a positive β such that

U − U_{n}^{(1)} ∈ W_{2,β}^{n+2} ∩ W̊_{2,β}^{2} (Λ ∩ {ξ > 2R})

(cf. [5], [8]). Hence and from the Sobolev embedding theorem it follows that U−U_{n}^{(1)} and its derivatives up to order n have the estimate O( exp(−βξ)) as ξ → +∞. Thus U = U_{n}^{(1)} + (U − U_{n}^{(1)}) admits the estimate ∇^{k}U(ξ, η) = O(|ξ|^{−n+1}) for k = 1, …, n. Therefore the Airy function F (z), equal to F_{1}(z) + F_{2}(z), has the representation

F (z) = O(|z|^{n−1}) and ∇^{k}F (z) = O(|z|^{n−1−2k}), k = 1, …, n .

A displacement vector v_{n}^{(2)} corresponding to F (z) has the form

v_{n}^{(2)}(z) = α + icz + v_{n0}^{(2)}(z),

where c is a real coefficient and ∇^{k}v_{n0}^{(2)}(z) = O(|z|^{n−2k−2}). Since the gradient ∇v_{n}^{(2)} of the displacement vector v_{n}^{(2)} is square summable in a neighborhood of infinity, we have

∫_{|z|=R_{n}} T v_{n}^{(2)}ds → 0

for a certain sequence {R_{n}}, R_{n} → ∞. This and the condition ∫_{S} h ds = 0 imply

∫_{S} T v_{n}^{(2)} ds = 0.

From the first Kolosov formula

ΔF (z) = 4 Re(φ_{n}^{(2)})′(z)

(see [3], Sect. 8.4) it follows that Re(φ_{n}^{(2)})′(z), where φ_{n}^{(2)} is the complex potential corresponding to v_{n}^{(2)}, is square summable in a neighborhood of infinity. Therefore

φ_{n}^{(2)}(z) = icz + ∑_{k=−∞}^{0} b_{k}z^{−k}, c ∈ R.

The second Kolosov formula

\frac{∂^{2}F}{∂x^{2}} (z) − \frac{∂^{2}F}{∂y^{2}} (z) − 2i\frac{∂^{2}F}{∂x∂y} (z) = 2[\overline{z}(φ_{n}^{(2)})′′(z) + (ψ_{n}^{(2)})′(z)]

(see [3], Sect. 8.4) implies that the derivative of the second complex potential ψ_{n}^{(2)}(z) is also square summable in a neighborhood of infinity. Therefore we have

ψ_{n}^{(2)}(z) = ∑_{k=−∞}^{0} a_{k}z^{−k}.

Thus the displacement vector v_{n}^{(2)} admits the representation

v_{n}^{(2)}(z) = \frac{1}{2μ} (κ + 1)icz + O(1).

Hence, up to a rigid displacement equal to (2μ)^{−1}(κ + 1)icz, the constructed displacement vector v_{n}^{(2)} is bounded at infinity. We remove this rigid displacement term since it is a solution of the homogeneous problem N^{−}. Thus the displacement vector v = v_{n} + v_{n}^{(1)} + v_{n}^{(2)} is a solution of the problem N^{−}. It has the required representation (13) and satisfies condition (12).

Corollary 3.1. Let the polynomials H_{±}^{(1)} satisfy

Im (H_{+}^{(1)}(0) + H_{−}^{(1)}(0)) = Im ((∂H_{+}^{(1)}/∂t)(0) + (∂H_{−}^{(1)}/∂t)(0)) = 0 .

Then the solution of the problem N^{−} has a finite energy integral and can be represented in form (13) with

φ_{n}(z) = i ( ∑_{m=0}^{p+1} β_{1,m} (log \frac{zz_{0}}{z_{0} − z} )^{m}) (\frac{zz_{0}}{z_{0} − z} )^{ν−1}

+ ∑_{k=2}^{n+1} P_{φ}^{(k+2)} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k+ν−2} ,

ψ_{n}(z) = i ( ∑_{m=0}^{p+1} γ_{1,m} (log \frac{zz_{0}}{z_{0} − z} )^{m}) (\frac{zz_{0}}{z_{0} − z} )^{ν−1}

+ ∑_{k=2}^{n+1} P_{ψ}^{(k+2)} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k+ν−2} ,

where β_{1,m}, γ_{1,m} are real numbers, p = 1 if ν = −1, 0, 1, 2, and p = 2 otherwise, and P_{φ}^{(j)}, P_{ψ}^{(j)} are polynomials of degree j.

Corollary 3.2. Let h have the representation

h_{±}(z) = h_{±}^{(1)}x^{−1/2} + h_{±}^{(2)}x^{1/2} log x + h_{±}^{(3)}x^{1/2} + …

on S_{±}. Then the problem N^{−} with the boundary data h has a solution in the class 𝔐 if and only if

Im (h_{+}^{(1)} + h_{−}^{(1)}) = 0, Im (h_{+}^{(2)} + h_{−}^{(2)}) = 0,

Re h_{+}^{(1)}α_{−} + Re h_{−}^{(1)}α_{+} − \frac{1}{2} Im (h_{+}^{(3)} + h_{−}^{(3)}) = 0 .

Theorem 4. Let Ω have an outward peak. Suppose that h is a C^{∞}-function on S\{O} and its restrictions to the arcs S_{±} have the representations

h_{±}(z) = ∑_{k=0}^{n−1} H_{±}^{(k)}(log x)x^{ν+k} + O(x^{ν+n}), ν > −2.

Here H_{±}^{(j)} are polynomials of degree at most j. Suppose the above representations can be differentiated n times and

V.P. ∫_{S} h ds = 0 .

Then the problem N^{−} with the boundary data h on S has a solution v bounded at infinity, admitting the representation

v(z) = \frac{1}{2μ}[κ (φ_{n}(z) + φ_{∗}(z)) − z\overline{(φ′_{n}(z) + φ′_{∗}(z))}
− (\overline{ψ_{n}(z) + ψ_{∗}(z)}) ] + v_{0}(z)  (18)

up to a linear function α+icz with real coefficient c, and satisfying the condition

∫_{S} T v ds = 0 .

Here ∇^{k}v_{0}(z) = O(|z|^{n+ν−k}), k = 0, …, n − 1, the complex potentials φ_{n}, ψ_{n} have the form

φ_{n}(z) = ∑_{k=1}^{n} P_{φ}^{(k+1)} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k+ν} ,

ψ_{n}(z) = ∑_{k=1}^{n} P_{ψ}^{(k+1)} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k+ν} ,

where P_{φ}^{(j)} and P_{ψ}^{(j)} are polynomials of degree j, and

φ_{∗}(z) = ∑_{k=1}^{m} R_{φ,k} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k/2} ,

ψ_{∗}(z) = ∑_{k=1}^{m} R_{ψ,k} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k/2} ,

where R_{φ,k} and R_{ψ,k} are polynomials of degree at most [(k − 1)/2] and m is the largest integer not exceeding 2(n + ν) + 1.

Proof. We choose a displacement vector v_{n} such that h_{n} = h − T v_{n} belongs to C^{∞}(S\{O}) and (h_{n})_{±}(z) = O(x^{n+ν}). To this end, as in Theorem 3, we use the method of complex stress functions. It is convenient to write the boundary conditions of the problem N^{−} in the Muskhelishvili form

φ(z) + z\overline{φ′(z)} + \overline{ψ(z)} = f(z), z ∈ S .

Here

f(z) = −i ∫_{(0z)^{⌣}} h ds + const ,

where by (0z)^{⌣} we denote the arc of S connecting 0 and z. It suffices to consider a function f(z) of the form ±ih_{±}x^{ν+1}(log x)^{m} on arcs S_{±}. We set

φ_{n}(z) = β_{m} (\frac{zz_{0}}{z − z_{0}} )^{ν+1} (log \frac{zz_{0}}{z − z_{0}} )^{m} ,

ψ_{n}(z) = γ_{m} (\frac{zz_{0}}{z − z_{0}} )^{ν+1} (log \frac{zz_{0}}{z − z_{0}} )^{m}

for ν ≠ n/2, n ∈ Z, where z_{0} is a fixed point in Ω.

We show that there exist β_{m} and γ_{m} such that the function φ_{n}(z)+z\overline{φ_{n} ′(z)}+ \overline{ψ_{n}(z)} on S_{±} is equal to the sum of ±ih_{±}x^{ν+1}(log x)^{m} and a linear combination of the functions c_{±}x^{i}(log x)^{j}, admitting the estimate O(x^{ν+1}(log x)^{m−1}).

As in Theorem 2, we decompose φ_{n} and ψ_{n} in powers of x along S_{±}. Coefficients β_{m} and γ_{m} in x^{ν+1}(log x)^{m} are found from the system

\{ \begin{matrix}β_{m} + (ν + 1)\overline{β_{m}} + \overline{γ_{m}} = ih_{+} \\ e^{4iπν}β_{m} + (ν + 1)\overline{β_{m}} + \overline{γ_{m}} = −ih_{−}e^{2iπν} . \end{matrix}

In the case ν = n/2, n ∈ Z the functions φ_{n} and ψ_{n} are defined by

φ_{n}(z) = (β_{m,1} (log \frac{zz_{0}}{z − z_{0}} )^{m+1} + β_{m,0} (log \frac{zz_{0}}{z − z_{0}} )^{m}) (\frac{zz_{0}}{z − z_{0}} )^{ν+1} ,

ψ_{n}(z) = (γ_{m,1} (log \frac{zz_{0}}{z − z_{0}} )^{m+1} + γ_{m,0} (log \frac{zz_{0}}{z − z_{0}} )^{m}) (\frac{zz_{0}}{z − z_{0}} )^{ν+1} ,

where z_{0} is a fixed point in Ω. The above coefficients β_{m,1} and γ_{m,1} are found from the system

\{ \begin{matrix}β_{m,1} + (ν + 1)\overline{β_{m,1}} + \overline{γ_{m,1}} = 0 , \\ β_{m,1} − (ν + 1)\overline{β_{m,1}} − \overline{γ_{m,1}} = −\frac{h_{+} + (−1)^{m}h_{−}}{2π(m + 1)} \end{matrix} .

Given β_{m,0}, we find γ_{m,0} by the equation

β_{m,0} + (ν + 1)\overline{β_{m,0}} + \overline{γ_{m,0}} = ih_{+} + (m + 1)\overline{β_{m,1}} .

After the complex stress functions φ_{n} and ψ_{n} have been found, the displacement vector v_{n} is defined by means of

v_{n}(z) = (2μ)^{−1} (κφ_{n}(z) − z\overline{φ ′_{n}(z)} − \overline{ψ_{n}(z)}) .

Since V.P. ∫_{S} h ds = 0, we have H_{+}^{(0)} = −H_{−}^{(0)} for −2 < ν ≤ −1. Therefore

lim_{ε→0} ∫_{{q∈Ω^{c}, |q|=ε}} T v_{n}ds = 0 .

Hence we arrive at V.P. ∫_{S} T v_{n}ds = 0.

(ii) As in Theorem 3, we can construct the displacement vector v_{n}^{(1)} so that

v_{n}^{(1)}(z) = O(|z|^{n+[ν]+1}) as z → 0

and v_{n}^{(1)} vanishes at infinity. Furthermore, the stress function h_{n}^{(1)} = h_{n} − T v_{n}^{(1)} on S has the zero principal vector and the zero principal moment with respect to any point z_{0} ∈ Ω.

We find a displacement vector v_{n}^{(2)} with given h_{n}^{(1)} on S. To this end, we express the boundary condition of this problem via the Airy function F in Ω^{c}. Let F = b and ∂F/∂n = d on S. Taking into account the relation of b and d with h_{n}^{(1)} we have d(z) = O(|z|^{n+1+ν}), b(z) = O(|z|^{n+2+ν}) for z ∈ S\{O}.

We represent the Airy function F (z) as the sum F_{1}(z) + F_{2}(z), where F_{1}(z) is chosen so that it vanishes at infinity, satisfies the conditions

F_{1}(z) = b(z), ∂F_{1}(z)/∂n = d(z) on S

and admits the estimate ∇^{k}F_{1}(z) = O(|z|^{n+2+ν−k}), k = 0, …, n + 2, in a neighborhood of the origin. Then F_{2}(z) is a unique solution of problem

△^{2}F_{2} = −△^{2}F_{1} in Ω^{c}, F_{2} = ∂F_{2}/∂n = 0 on S

in W̊^{2,ρ}(Ω^{c}) with ρ(z) = (1 + |z|)^{2} (see [4]).

We make the change of variable t = log(1/r), r = |z|. Let Λ be the image of the domain Ω^{c} under the mapping (r, θ) → (t, θ), where r, θ are polar coordinates of (x, y). The function U(t, θ) = F_{2}(e^{−t}, θ) solves the equation

L(∂_{t}, ∂_{θ})U(t, θ) = [(\frac{∂}{∂t} + 2)^{2} + \frac{∂^{2}}{∂θ^{2}} ] [\frac{∂^{2}}{∂t^{2}} + \frac{∂^{2}}{∂θ^{2}} ] U(t, θ)
= H(t, θ) in Λ ∩ {t > R}

and satisfies U = ∂U/∂n = 0 on ∂Λ∩{t > R}, where ∇^{k}H(t, θ) = O(e^{−(n+2+ν)t}), k = 0, …, n − 2. From the local estimate

∥U∥_{W_{2}^{n+2}(Λ∩{ℓ−1<ξ<ℓ+1})} ≤ const(∥χH∥_{W_{2}^{n−2}(Λ)} + ∥χU∥_{L_{2}(Λ)}) ,

where χ belongs to C_{0}^{∞}(ℓ − 2, ℓ + 2) and equals 1 in (ℓ − 1, ℓ + 1), it follows that U belongs to W̊_{2,1}^{2} ∩ W_{2,1}^{n+2}(Λ ∩ {t > R}). By U(∂_{t}, ∂_{θ}) we denote the operator of the boundary value problem

U(∂_{t}, ∂_{θ})U = F in Π = {(t, θ) : 0 < θ < 2π}, U = (∂/∂n)U = 0 on ∂Π

continuously mapping W_{2}^{n+2} ∩ W̊_{2}^{2}(Π) into W_{2}^{n−2}(Π). Let β satisfy n + ν + 1 < β < n + ν + 3/2. The operator pencil U(ik, ∂_{θ}) : W_{2}^{n+2} ∩ W̊_{2}^{2}(0, 2π) → W_{2}^{n−2}(0, 2π) has p = 2n+[2ν]+2 eigenvalues of the form k = iℓ/2, ℓ ∈ N, in the strip {k : 0 < Im k < β}. The eigenvalues have multiplicity 2 and multiplicity of each eigenvector equals 1. Therefore the solution U admits the representation (cf. [5], [8])

U = ∑_{k=1}^{p} c_{k}U_{k} + W.

Here U_{k} are linearly independent, each U_{k} satisfies the equation LU_{k} = 0 in the domain Λ ∩ {t > R}, vanishes on ∂Λ ∩ {t > R} and U_{k} ∉ W_{2,β}^{n+2}(Λ ∩ {t > R}), in addition, W ∈ W_{2,β}^{n+2}(Λ ∩ {t > R}).

Making the inverse change of variable we obtain

F (r, θ) = ∑_{k=1}^{p} c_{k}U_{k}(− log r, θ) + O(r^{β})

and this equality can be differentiated n times. By the method of complex stress functions the displacement vector v_{n}^{(2)} corresponding to F is given by

v_{n}^{(2)}(z) = \frac{1}{2μ} [κφ_{∗}(z) − z\overline{φ′_{∗}(z)} − \overline{ψ_{∗}(z)}] + w^{(1)}(z) ,

where w^{(1)}(z) = α + icz + O(|z|^{n+ν}) with real α and this equality can be differentiated n − 1 times,

φ_{∗}(z) = ∑_{k=1}^{m} R_{φ,k} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k/2}

and

ψ_{∗}(z) = ∑_{k=1}^{m} R_{ψ,k} (log \frac{zz_{0}}{z_{0} − z} ) (\frac{zz_{0}}{z_{0} − z} )^{k/2} .

Here m is the largest integer not exceeding 2(n + ν) + 1. As in Theorem 3, we can prove that

∫_{S} T v_{n}^{(2)}ds = 0.

Hence the displacement vector v = v_{n} +v_{n}^{(1)} +v_{n}^{(2)} is a solution of the problem N^{−} with the boundary data h and admits the required representation.

Corollary 4.1. Let h have representations

h_{±}(z) = ∑_{k=0}^{n−1} h_{±}^{(k)}x^{k+ν} + O(x^{n+ν}), x ∈ S_{±},

for ν > −1, ν ≠ ℓ/2, ℓ ∈ Z. Then the functions φ_{n} and ψ_{n} in (18) have the form

φ_{n}(z) = ∑_{k=1}^{n} β_{k} (\frac{zz_{0}}{z − z_{0}} )^{k+ν} , ψ_{n}(z) = ∑_{k=1}^{n} γ_{k} (\frac{zz_{0}}{z − z_{0}} )^{k+ν} .

2.3. Properties of solutions of the problem D^{−}.

Proposition 1. Let Ω have an outward peak and let a vector-valued function g satisfy on S_{±} the conditions

\frac{∂^{k}g}{∂s^{k}} (z) = O(|z|^{β−k}), k = 0, 1, 2.

Then the problem D^{−} with the boundary data g has a solution u bounded at infinity and subject to

∫_{S} |q||∇u(q)|ds_{q} < +∞, u(z) = O(|z|^{γ})  (19)

if −1 < γ < β ≤ 0, and to

∫_{S} |∇u(q)|ds_{q} < +∞, u(z) = O(|z|^{γ})  (20)

if 0 < γ < min{β, 1/2}.

Proof. As in Theorem 2, we can construct a solution u satisfying either (19) or (20) and such that u belongs to W^{1,ρ}(|z| > R) and is square summable in a neighborhood of infinity for sufficiently large R. We check now that u is bounded at infinity. By the first Kolosov formula Re{φ′(z)} is a linear combination of the components of ∇u. It follows that Re φ′(z) is square summable in a neighborhood of infinity. So, φ(z) has the decomposition

φ(z) = icz + ∑_{k=−∞}^{0} b_{k}z^{−k}, c ∈ R.

From the equality

u(z) = (2μ)^{−1}(κφ(z) − z\overline{φ′(z)} − \overline{ψ(z)})

it follows that ψ(z) is also square summable in a neighborhood of infinity. Therefore ψ(z) can be presented in the form

ψ(z) = ∑_{k=−∞}^{0} d_{k}z^{−k}.

Finally we have

u(z) = i(2μ)^{−1}(κ + 1)cz + O(1).

Since the function ρ^{−1}u^{−} is square summable in a neighborhood of infinity, c = 0. Consequently, u is bounded at infinity.

Proposition 2. Let Ω have an inward peak and let g on S_{±} satisfy the conditions

\frac{∂^{k}g}{∂s^{k}} (z) = O(|z|^{β−k}), k = 0, 1, 2.

Then the problem D^{−} with the boundary data g has a solution u bounded at infinity and satisfying

∫_{S} |u(q)|ds_{q} < +∞, ∫_{S} |q||∇u(q)|ds_{q} < +∞.  (21)

Proof. Let u^{(1)} be an extension of g onto Ω^{c} equal to zero outside a certain disk and satisfying Δ^{∗}u^{(1)} = O(|z|^{β−3}) as z → 0. We look for a function u^{(2)} such that

△^{∗}u^{(2)} = −△^{∗}u^{(1)} in Ω^{c}, u^{(2)} = 0 on S\{O} .  (22)

After the change of variables z = 1/ζ (ζ = ξ + iη) the equation in (22) takes the form

Δ^{∗}U + L(∂_{ξ}, ∂_{η})U = F in Λ,  (23)

where U(ξ, η) = u^{(2)}(ξ/|ζ|^{2}, η/|ζ|^{2}), Λ is the image of Ω and L(∂_{ξ}, ∂_{η}) is the second order differential operator with coefficients admitting the estimate O(ξ^{−β−1}). Since F ∈ L_{2}(Λ), problem (23) has a unique solution in the space W_{2}^{2}(Λ). Hence u = u^{(1)} + u^{(2)} satisfies conditions (21).

3. Integral Equation on the Contour with an Inward Peak

3.1. Integral equation for the problem D^{+}. We shall use the following lemma.

Lemma 1. Let Ω have an inward peak and let α = max(|κ+^{′′}(0)|, |κ−^{′′}(0)|). Then, for any σ = (σ^{(1)}, σ^{(2)}) from 𝔐,

(W σ^{(1)}) (z)=∓\frac{1}{2} (σ_{+}^{(1)}−σ_{−}^{(1)}) ± \frac{μ}{2π(λ + 2μ)} ∫_{0}^{δ}(σ_{+}^{(2)}−σ_{−}^{(2)}) (τ) \frac{dτ}{x − τ} +O(1),

(W σ^{(2)}) (z)=∓ \frac{1}{2} (σ_{+}^{(2)}−σ_{−}^{(2)}) ∓ \frac{μ}{2π(λ + 2μ)} ∫_{0}^{δ}(σ_{+}^{(1)}−σ_{−}^{(1)}) (τ) \frac{dτ}{x − τ} +O(1)

with the upper sign if

z ∈ {(x, y) : κ_{+}(x) < y < αx, x ∈ (0, δ)} ,

and with the lower sign if

z ∈ {(x, y) : −αx^{2} < y < κ_{−}(x), x ∈ (0, δ)} .

For the proof see [9]. Now we state a uniqueness theorem.

Theorem 5. Let Ω have an inward peak. The boundary integral equation

2^{−1}σ − W σ + c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3} = 0  (24)

has only the trivial solution in the Cartesian 𝔐 × R^{3}.

Proof. Let a pair (σ, c) with σ ∈ 𝔐, c ∈ R^{3}, be a solution of equation (24). Consider the displacement vector

W σ(z) + c_{1}A_{1}(z) + c_{2}A_{2}(z) + c_{3}A_{3}(z), z ∈ Ω,  (25)

with zero boundary value on S\{O}. We show that this vector-valued function is equal to zero in Ω.

We use the change of the variable t = log 1/r and denote by W_{1}(z), W_{2}(z) the components of (25). Then U(t, θ) with the components

W_{1}(e^{−t}, θ) cos θ + W_{2}(e^{−t}, θ) sin θ, −W_{1}(e^{−t}, θ) sin θ + W_{2}(e^{−t}, θ) cos θ

is a solution of the problem

L(∂_{t}, ∂_{θ})U = (△^{∗} + K)U = 0 in Λ, U = 0 on ∂Λ,

where r, θ are polar coordinates of (x, y) and Λ is the image of Ω. Here by K we denote the first order differential operator with constant coefficients

K = (\begin{matrix}−λ + 2μ, & −(λ + 3μ)∂/∂θ \\ (λ + 3μ)∂/∂θ & −μ \end{matrix}) .

Since the potential W σ(z) grows not faster than a power function as z → 0, there exists β < 0 such that U ∈ L_{2}(Λ). By the local estimate (11) we obtain that U belongs to W_{2,β}^{2} ∩ W̊_{2,β}^{1}(Λ). The eigenvalues of the operator pencil D(ik, ∂_{θ}) : W_{2}^{2} ∩ W̊_{2}^{1}(0, 2π) → L_{2}(0, 2π) are the numbers iℓ/2, where ℓ is a nonvanishing integer. According to [5], [8], the operator L is invertible if β ≠ iℓ/2, ℓ = ±1, ±2, . . .. Since U ∈ ker L, from its asymptotic representation of U (cf. [5]) it follows that either U(t, θ) = O(e^{−t/2}) or U(t, θ) grows faster than e^{t/(2+ε)} (ε > 0) as t → +∞. In the first case the energy integral of (W_{1}, W_{2}) is finite and therefore it is equal to zero in Ω. Now we shall show that the second case is impossible.

Let χ be a C^{∞}-function equal to zero outside of a small neighborhood of the origin and χ = 1 near the origin. The components of χσ will be denoted by σ^{(1)}, σ^{(2)}. Since the boundary values of W σ are bounded on S\{O}, by Lemma 1 we have

− \frac{1}{2} (σ_{+}^{(1)} − σ_{−}^{(1)}) (x) + \frac{μ}{2π(λ + 2μ)} ∫_{0}^{δ} (σ_{+}^{(2)} − σ_{−}^{(2)}) (τ) \frac{dτ}{x − τ} = O(1),

− \frac{1}{2} (σ_{+}^{(2)} − σ_{−}^{(2)}) (x) − \frac{μ}{2π(λ + 2μ)} ∫_{0}^{δ} (σ_{+}^{(1)} − σ_{−}^{(1)}) (τ) \frac{dτ}{x − τ} = O(1).

These equalities imply that the functions (σ_{+}^{(1)} − σ_{−}^{(1)}) and (σ_{+}^{(2)} − σ_{−}^{(2)}) are psummable for any p > 1 in a neighborhood of the origin. Therefore W σ(z) is also p-summable for any p > 1 along any ray

{(x, y) : x = αt, y = βt, t > 0}

near the origin with α < 0. Since A_{j} = O(1), j = 1, 2, 3, our goal is achieved. Hence we have

W σ(z) + c_{1}A_{1}(z) + c_{2}A_{2}(z) + c_{3}A_{3}(z) = 0 in Ω.

Let u^{−} be a solution of the problem D^{−} with the boundary data σ. We notice that

V T u^{−} − W u^{−} = u^{−}(∞) in Ω .

Here by V τ(z) we denote the simple-layer potential with a density τ and with the kernel

Γ(z, q) − Γ(z, 0), q ∈ S\{O}.

Since

W u^{−} = − ∑_{k=1}^{3} c_{k}A_{k} ,

by substituting W u^{−} into the previous equality, we find

V T u^{−} + ∑_{k=1}^{3} c_{k}A_{k} = u^{−}(∞) in Ω.

The limit relations for the simple-layer potential imply

T (V T u^{−} + ∑_{k=1}^{3} c_{k}A_{k} + \frac{1}{2}u^{−}) = 0 on S\\{O\}.

Let w_{k}^{−} be the solution of the problem

△^{∗}w_{k}^{−} = 0 in Ω^{c}, T w_{k}^{−} = T A_{k} on S\{O}

constructed in Theorem 3. Since the boundary function T A_{k} does not satisfy the conditions of Corollary 3.2, it follows that the solution w_{k}^{−} does not belong to 𝔐. Then we have

V T u^{−}(z) = −u^{−}(z) − ∑_{k=1}^{3} c_{k}w_{k}^{−}(z) + u_{0}(z) in Ω^{c},  (26)

where u_{0} is the displacement vector satisfying T u_{0} = 0 on S\{O}. We substitute (26) into the identity

W u^{−} − V T u^{−} = u^{−} − u^{−}(∞) in Ω^{c}

and obtain

W σ(z) = u_{0}(z) − ∑_{k=1}^{3} c_{k}w_{k}^{−} − u^{−}(∞), z ∈ Ω^{c} .  (27)

From the jump formula for W σ it follows

σ(p) = −u_{0}(z) + ∑_{k=1}^{3} c_{k}w_{k}^{−} + u^{−}(∞) − ∑_{k=1}^{3} c_{k}A_{k}.

Since the density σ belongs to 𝔐, according to (27) the function w_{0}, defined by w_{0}(ζ) = u_{0}(ζ^{−1}), and ∇w_{0} grow not faster than a power function in the image Λ of Ω^{c}. Therefore w_{0} ∈ W_{2,β}^{2}(Λ) with a small negative β. Since ∇u_{0}(z) = O(|z|^{−2}) as z → ∞ and T u_{0} = 0 on S\{O}, it follows from Betti’s formula that

∫_{{κ_{−}(x)<y<κ_{+}(x)}} T u_{0} ds = 0 for 0 < x < δ .

We represent the displacement vector u_{0} via the Airy function U in the domain Ω_{δ} = Ω^{c} ∩ {|z| < δ}. Since

\frac{∂U}{∂x} + i\frac{∂U}{∂y} = −i ∫_{(0z)^{⌣}} T u_{0} ds + const ,

where by (0z)^{⌣} we denote the arc of S connecting 0 and z, and since (∂U/∂x)+ i(∂U/∂y) is defined up to a constant term, we assume that

\frac{∂U}{∂x} + i\frac{∂U}{∂y} = 0 on S_{±}.

Therefore U is equal to the constants c_{±} on S_{±}. We use the Kelvin transform

W (ζ) = |ζ|^{2}U(1/ζ), ζ ∈ Λ_{δ},

where Λ_{δ} is the image Ω_{δ} under the mapping ζ = 1/z. The expressions W (ζ) and ∇^{k}W (ζ), k = 1, 2, grow not faster than a power function. Therefore, W belongs to W_{2,β}^{2}(Λ_{δ}) with β < 0, W is biharmonic in Λ_{δ} and satisfies the conditions

W (ζ) = c_{±}|ζ|^{2}, (∂W/∂n) = c_{±}(∂|ζ|^{2}/∂n) on ∂Λ_{δ} ∩ {|ζ| > δ^{−1}}.

From the local estimate

∥W ∥_{W_{2}^{3}(Λ_{1}∩{ℓ−1<Reζ<ℓ+1})}

≤ const[∥χW ∥_{W_{2}^{5/2}(∂Λ_{δ})} + ∥χ(∂W/∂n)∥_{W_{2}^{3/2}(∂Λ_{δ})} + ∥χW ∥_{W_{2}^{2}(Λ_{δ})}], (28)

where a cut-off C^{∞} function χ equals 1 on (ℓ − 1, ℓ + 1) and vanishes outside (ℓ − 2, ℓ + 2), it follows that W belongs to W_{2,β}^{3}(Λ_{δ}).

Since ∇^{2}w_{0} is represented as a linear combination of derivatives up to the third order of W with coefficients growing not faster than a power function, we have w_{0} ∈ W_{2,β}^{3}(Λ_{δ}).

Let N (∂_{ξ}, ∂_{η}) denote the operator of the boundary value problem

△^{∗}u = f in Π = {(ξ, η) : ξ ∈ R^{1}, −κ′′_{+}(0)/2 < η < −κ′′_{−}(0)/2} ,
T u = h on {(ξ, η) : η = −κ′′_{±}(0)/2} ,

continuously mapping W_{2,β}^{2}(Π) into L_{2,β}(Π) × W_{2,β}^{1/2}(∂Π). It is well known that the eigenvalues of the operator pencil N (ik, ∂_{η}): W_{2}^{2}(−κ′′_{+}(0)/2, −κ′′_{−}(0)/2) → L_{2}(−κ′′_{+}(0)/2, −κ′′_{−}(0)/2) × R^{2} are found from the equation

k^{2}α^{2} = (sinh k α)^{2} ,

where α = (κ′′_{+}(0) − κ′′_{−}(0)) /2. One can check that the dimension of the space

Ker N (0, D_{η}) equals two. The vector-valued function u_{1}^{(0)}(η) = (0, 1) is the basis element of multiplicity 4. The second linear independent eigenvector u_{2}^{(0)}(η) = (1, 0) has multiplicity 2. The generalized eigenvectors corresponding to u_{1}^{(0)}, u_{2}^{(0)} are

u_{1}^{(1)}(η) = (−iη − \frac{1}{2} i (α_{+} + α_{−}), 0) ,
u_{1}^{(2)}(η) = (0, −\frac{1}{2} \frac{λ}{λ + 2μ} η^{2} − \frac{1}{2} \frac{λ}{λ + 2μ} (α_{+} + α_{−})η) ,
u_{1}^{(3)}(η) = ( − \frac{i}{6} \frac{3λ + 4μ}{λ + 2μ} η^{3} − \frac{i}{4}(α_{+} + α_{−})\frac{3λ + 4μ}{λ + 2μ} η^{2} −

− 2i \frac{λ + μ}{λ + 2μ} α_{+}α_{−}η, 0),

u_{2}^{(1)}(η) = (0, −i \frac{λ}{λ + 2μ} η) .

The operator pencil N (ik, ∂_{η}) has no other eigenvalues on R.

Therefore u_{0}(z) is represented near the origin in the form

u_{0}(z) = ∑_{k=0}^{6} d_{k}Z_{k}(z) + u_{1}(z)

(cf. [8]). Here Z_{k} are linearly independent solutions of the equation Δ^{∗}u = 0 in a vicinity of the pick in Ω^{c} satisfying the boundary condition T u = 0 near O on S. The last term u_{1} and its gradient exponentially vanish as z → 0. Three of the vector-valued functions Z_{k}, 1 ≤ k ≤ 6, say Z_{1}, Z_{2}, Z_{3}, form rigid displacements. The others are represented in the form

Z_{4}(z) = x^{−1} + O(1), Z_{5}(z) = ix^{−3} + O(x^{−2}),
Z_{6}(z) = ix^{−4} + O(x^{−3}).

Since the functions w_{1}^{−}, w_{2}^{−}, w_{3}^{−}, Z_{4}, Z_{5}, Z_{6} have different orders of singularities and since σ ∈ 𝔐, we have

c_{1} = c_{2} = c_{3} = d_{4} = d_{5} = d_{6} = 0.

In particular,

u_{0}(z) = d_{1}Z_{1}(z) + d_{2}Z_{2}(z) + d_{3}Z_{3}(z) + u_{1}(z).

Hence it follows that u_{0} and its gradient ∇u_{0} are bounded as z → 0. On the other hand, according to (25) we have

u_{0}(z) − u_{0}(∞) = O(|z|^{−1}), ∇u_{0}(z) = O(|z|^{−2}) as z → ∞.

By the classical uniqueness theorem (cf. [16], Ch. I, Sect. 20) it follows that u_{0}(z) = u_{0}(∞), z ∈ Ω^{c}. Thus

σ(p) = −u_{0}(p) + u^{−}(∞) = const on S\{O}.

Since the nonzero constant does not satisfy the homogeneous equation (24), it follows that σ = 0.

Consider the equation

2^{−1}σ − W σ − c_{1}A_{1} − c_{2}A_{2} − c_{3}A_{3} = −g  (29)

on S with respect to a density σ and a vector (c_{1}, c_{2}, c_{3}). The functions A_{1}, A_{2}, A_{3} have been defined in the Introduction.

Theorem 6. Let Ω have an inward peak and let g belong to the class 𝔑_{ν}, ν > 3. Then equation (29) has a unique solution {σ, c_{1}, c_{2}, c_{3}} in 𝔐 × R^{3}, and the density σ can be represented in the form

σ(z) = (α(log x)^{2} + β log x + γ) x^{−1/2} + O(x^{−ε}) on S_{±} ,

with small positive ε.

Proof. Let U_{m,k}, m = 1, …, 4, and k = 1, 2, denote the solutions of the problem

△^{∗}u = 0 in Ω ∩ B_{r}, u = 0 on B_{r} ∩ S\{O}

corresponding to the eigenvalues k = iℓ/2, ℓ ∈ Z, ℓ ≠ 0, of the operator pencil introduced in Theorem 2. Here B_{r} = {|z| < r} with small positive r. We normalize the functions U_{m,k} in the following way:

U_{11}(z) = \frac{i}{2μ} [2κ Im z^{1/2} − \overline{z^{−1/2}} Im z] +

+ i \frac{(κ − 1)(α_{+} − α_{−})}{8πκμ}[2κ Im (z^{3/2} log z) − 3\overline{z^{1/2} log z} Im z −

− 2\overline{z^{1/2}} Im z] − i \frac{(κ − 1)α_{+}}{2μ} \overline{z^{3/2}} + O(z^{5/2}(log z)^{2}),

U_{12}(z) = −\frac{1}{2μ} [2κ Im z^{1/2} + \overline{z^{−1/2}} Im z] −

− \frac{(κ + 1)(α_{+} − α_{−})}{8πκμ}[2κ Im (z^{3/2} log z) + 3\overline{z^{1/2} log z} Im z +

+ 2\overline{z^{1/2}} Im z] + \frac{(κ + 1)α_{+}}{2μ} \overline{z^{3/2}} + O(z^{5/2}(log z)^{2}),

U_{21}(z) = i \frac{κ − 1}{μ} Im z + i \frac{(κ − 1)(α_{+} − α_{−})}{4πμκ} [2κ Im (z^{2} log z) −

− 4\overline{z log z} Im z − 2\overline{z} Im z] − i \frac{(κ − 1)α_{+}}{μ} \overline{z^{2}} + O(z^{3} log^{2} z),

U_{22}(z) = −\frac{κ + 1}{μ} Im z − \frac{(α_{+} − α_{−})(κ + 1)}{4πμκ}[2κ Im (z^{2} log z) +

+ 4\overline{z log z} Im z + 2\overline{z} Im z] + \frac{(κ + 1)α_{+}}{μ} \overline{z^{2}} + O(z^{3}(log z)^{2}),

U_{31}(z) = \frac{i}{2μ} [2κ Im z^{3/2} − 3\overline{z^{1/2}} Im z] + O(z^{5/2} log z),

U_{32}(z) = −\frac{1}{2μ} [2κ Im z^{3/2} + 3\overline{z^{1/2}} Im z] + O(z^{5/2} log z),

U_{41}(z) = \frac{i}{μ} [κ Im z^{2} − 2\overline{z} Im z] + O(z^{3} log z),

U_{42}(z) = −\frac{1}{μ} [κ Im z^{2} + 2\overline{z} Im z] + O(z^{3} log z),

where α_{±} = κ′′_{±}(0)/2. According to Theorem 2, the solution of the problem D^{+} with the boundary data g admits the representation

u^{+}(z) = ∑_{m=1}^{4} ∑_{k=1}^{2} d_{mk}U_{mk}(z) + W (z)

(cf. [5], [8]), where coefficients d_{m,k} are defined uniquely and

W (z) = O(|z|^{2+ε}), ε > 0.

We set

c_{1} = d_{11}, c_{2} = (Qd_{12} + d_{31})/(Q^{2} + 1), c_{3} = d_{22}.

Then the displacement vector

u^{+}(z) − ∑_{k=1}^{3} c_{k}A_{k}(z)

is the sum of a linear combination of the functions

U_{12}(z) − QU_{31}(z), U_{21}(z), U_{32}(z), U_{41}(z), U_{42}(z)

and the function W_{1}(z) admitting the estimate O(|z|^{2+ε}). According to Theorem 3 (see also Corollaries 3.1 and 3.2), the problem N^{−} with the boundary data

T (u^{+} − ∑_{k=1}^{3} c_{k}A_{k})

has a solution whose trace on S\{O} is in the class 𝔐.

Now, when the vector c = (c_{1}, c_{2}, c_{3}) is chosen, by U^{+} and U^{−} we denote solutions of the problems D^{+} and D^{−} with the data

g − ∑_{k=1}^{3} c_{k}A_{k} on S\{O}.

Then T U^{+} admits the estimate O (x^{−1/2}) on S\{O}. According to Proposition 2 the displacement vector U^{−}(z) is bounded at infinity and

∫_{S} |T U^{−}(p)||p|ds_{p} < +∞ .

Hence (cf. [7], Ch. 5, Sect. 1) we obtain

g(p) − ∑_{k=1}^{3} c_{k}A_{k}(p) = V (T U^{+} − T U^{−}) (p) + U^{e}(∞), p ∈ S\{O}.

Let V^{−} be a solution of the problem N^{−}:

V^{−} = 0 in Ω^{c}, T V^{−} = T U^{+} on S\{O}, V^{−}(∞) = 0 .

Applying Betti’s formula to w = V^{−} − U^{−} + U^{−}(∞) in Ω^{c} and using the jump relation for W w we obtain

w(p) − 2W w(p) = − 2V (T U^{+} − T U^{−}) =

= −2g(p) + 2 ∑_{k=1}^{3} c_{k}A_{k}(p) + 2U^{−}(∞), p ∈ S\{O}.

Hence the function

σ(p) = w(p) − U^{−}(∞) = V^{−}(p) − g(p) + ∑_{k=1}^{3} c_{k}A_{k}(p), p ∈ S\{O}

with chosen c_{1}, c_{2}, c_{3} is a solution of equation (29) in the class 𝔐. According to Theorems 2 and 3 σ has the required asymptotics and according to Theorem 5 the obtained solution (σ, c) is unique.

3.2. Integral equation for the problem N^{−}. We say that a function v(z), z ∈ Ω, belongs to the space W̊_{2}^{1} outside the peak if for any C_{0}^{∞}-function κ such that O ∉ supp κ one has κv ∈ W̊_{2}^{1}(Ω).

Lemma 2. Let Ω have an outward peak. If a vector-valued function v(z), z ∈ Ω, belongs to W̊_{2}^{1} outside the peak, satisfies △^{∗}v = 0 in a vicinity of the peak, and v(z) and ∇v(z) grow not faster than a power function, then there exists β < 0 such that v(z) and ∇v(z) admit the estimate O( exp(β Re \frac{1}{z})) as z → 0.

Proof. The function w defined by w(ζ) = (κv)(ζ^{−1}) satisfies (△^{∗} + L)w = 0 in Λ = {ζ : ζ^{−1} ∈ Ω} and w = 0 on ∂Λ. Here κ belongs to C_{0}^{∞}(R^{2}) and is equal to 1 in a neighborhood of the peak and L is a second order differential operator with coefficients vanishing as ζ → ∞. For small negative β we have w ∈ W̊_{2,β}^{1}(Λ). Hence and by the local estimate (9) it follows that w ∈ W_{2,β}^{3} ∩ W_{2,β}^{1}(Λ).

The operator D(∂_{ξ}, ∂_{η}) of the boundary value problem

△^{∗}u = f in Π = {(ξ, η) : −κ′′_{+}(0)/2 < η < −κ′′_{−}(0)/2} , u = 0 on ∂Π

continuously maps W_{2,β}^{2} ∩W̊_{2,β}^{1}(Π) into L_{2,β}(Π). The eigenvalues of the operator pencil

D(ik, ∂_{η}) : W_{2}^{2} ∩ W̊_{2}^{1}(−κ′′_{+}(0)/2, −κ′′_{−}(0)/2) → L_{2}(−κ′′_{+}(0)/2, −κ′′_{−}(0)/2) .

are nonzero solutions of the equation

α^{2}k^{2} = κ (sinh k α)^{2}

where α = (κ′′_{+}(0) − κ′′_{−}(0))/2 and κ = (λ + 3μ)/(λ + μ). We assume that there are no eigenvalues in the strip {k : β < Im k < −β}. Therefore w belongs to W_{2,−β}^{3} ∩ W̊_{2,−β}^{1}(Λ) (cf. [5], [8]). The Sobolev embedding theorem implies that w and ∇w admit the estimate O( exp(βξ)) as ξ → ∞.

Theorem 7. Let Ω have an inward peak. Then the homogeneous integral equation

−2^{−1}τ + T V τ = 0  (30)

has only the trivial solution in the class 𝔐.

Proof. Let τ ∈ 𝔐 satisfy (30). Integrating this equation over S we obtain

∫_{S} τ ds = 0.

Hence the potential v(z) = V τ(z) is a solution of the problem N^{+} with the boundary data τ, that is

T v(p) = τ(p), p ∈ S\{O}.

By the Betti integral representation for v we obtain W v = 0 in Ω. Therefore the density v of this double-layer potential is a solution of the homogeneous integral equation of the problem D^{+}

(1 − 2W) v(p) = 0, p ∈ S\{O}.

Since the restriction of v to S\{O} belongs to 𝔐, it follows by Theorem 5 that

v(p) = 0 for p ∈ S\{O}.

Since v(z) and ∇v(z) vanish at infinity and grow not faster than a power function as z → 0, Lemma 2 implies that v(z) = O( exp(β Re \frac{1}{z})) with negative β. According to the classical uniqueness theorem we have v = 0 in Ω^{c} (cf. [16]).

We make the change of the variable t = log 1/r and denote by v^{(1)}(z), v^{(2)}(z) the components of the displacement vector v(z), z ∈ Ω. Then the vector-valued function U(t, θ) = v(e^{−t}, θ)e^{−it}, where r, θ are the polar coordinates of (x, y), is a solution of the equation (△^{∗} + K)U = 0 in the image Λ of Ω satisfying U = 0 at ∂Λ. Here K is the first order differential operator

K = (\begin{matrix}−λ + 2μ, & −(λ + 3μ)∂/∂θ \\ (λ + 3μ)∂/∂θ & −μ \end{matrix}) .

The vector-valued function v belongs to W̊_{2,β}^{1}(Λ) with negative β. From the local estimate (11) it follows that v is an element of W_{2,β}^{2}(Λ). Consider the operator D(∂_{t}, ∂_{θ}) : W_{2,β}^{2} ∩ W̊_{2,β}^{1}(Π) → L_{2,β}(Π) of the boundary value problem

(△^{∗} + K)U = F in Π = {(t, θ) : 0 < θ < 2π, t ∈ R}, U = 0 on ∂Π,

and introduce the operator pencil D(ik, ∂_{θ}) : k ∈ C, continuously mapping W_{2}^{2} ∩W̊_{2}^{1}(0, 2π) into L_{2}(0, 2π) and considered in Theorem 2. Then the displacement vector v(z) is represented as a linear combination of linearly independent nonzero solutions of the homogeneous problem D^{+} (cf. [8]). Since the displacement vector v(z), z ∈ Ω, admits the estimate O(|z|^{β}), β > −1, one has

v(z) = d_{1}ζ_{1}(z) + d_{2}ζ_{2}(z),

where ζ_{1} and ζ_{2} are solutions of the homogeneous problem D^{+} admitting the estimate O(|z|^{−1/2}).

From the limit relations for V τ it follows

τ(p) = d_{1}T ζ_{1}(p) + d_{2}T ζ_{2}, p ∈ S\{O}.

Since the stresses T ζ_{k}, k = 1, 2, do not belong to 𝔐 and are of different order and since τ ∈ 𝔐, the coefficients d_{1} and d_{2} are zero.

Theorem 8. Let Ω have an inward peak and let h belong to 𝔑_{ν}, ν > 3, and satisfy

∫_{S} hds = 0.

Then the integral equation

−2^{−1}τ + T V τ = h

has a unique solution in the class 𝔐. The solution τ can be represented as

τ(z) = α_{±}x^{−1/2} + O(1).

Proof. Let v^{−} denote a solution of the problem N^{−} with the boundary data h on S, vanishing at infinity. According to Theorem 3, v^{−} has the representation

v^{−}(z) = (α^{(0)} + α^{(1)} log x)x^{ν−2} + O(x^{ν−1−ε}) on S.

By Theorem 2 the displacement vector u^{+}, equal to v^{−} on S\{O}, is given by

u^{+}(z) = \frac{1}{2μ} [κφ(z) − z\overline{φ′(z)} − \overline{ψ(z)}] ,

where φ(z) and ψ(z) admit the representations

φ(z) = βz^{1/2} + O(z), ψ(z) = γz^{1/2} + O(z).

Therefore we have

T u^{i}(z) = α_{±}x^{−1/2} + O(1) on S\{O}.

Hence

v^{−} = V (T u^{+} − h) on S\{O}.

Let us show that the same equality is valid in Ω^{c}. We put

v(z) = v^{−}(z) − V (T u^{+} − h)(z), z ∈ Ω^{c}.

As in Theorem 2, we can prove that ∇v(z) admits the estimate O(|z|^{−2}) at infinity. Hence and by Lemma 2 it follows that v belongs to W̊^{1,ρ}(Ω^{c}) with ρ(z) = (1 + |z|)^{2}. Since the problem D^{−} is uniquely solvable in W̊^{1,ρ}(Ω^{c}) (cf. [4]), v vanishes in Ω^{c}. So the density

τ = T u^{+} − h

satisfies the integral equation of the problem N^{−}. According to Theorem 7 the solution in the class 𝔐 is unique.

4. Integral Equations on the Contour with an Outward Peak

4.1. Integral equation for the problem D^{+}.

Theorem 9. Let Ω have an outward peak. The homogeneous integral equation (1) has a two-dimensional space of solutions in 𝔐.

Proof. By Theorem 4 the homogeneous problem N^{−} in the class of functions admitting the estimate O(|z|^{β}), β > −1, has a nonzero solution vanishing at infinity and satisfying

ζ(z) = cz^{−1/2} + O(1).

Furthermore, the function ζ span a two-dimensional real linear space. From the integral representation

ζ(z) = (W ζ)(z) − (V T ζ)(z), z ∈ Ω^{c},

by the jump formula for W ζ we get

ζ(p) − 2(W ζ)(p) = −2(V T ζ)(p) = 0, p ∈ S\{O}.

Thus, ζ(p) is a solution of the homogeneous integral equation

(1 − 2W) σ = 0  (31)

for the problem D^{+}.

Now, let σ ∈ 𝔐 be a non-trivial solution of equation (31). The potential W σ and ∇W σ in Ω have a power growth as z → 0. According to Lemma 2, W σ(z) = O( exp(β Re \frac{1}{z})) with negative β and therefore W σ ∈ W̊_{2}^{1}(Ω). Since the homogeneous problem D^{+} has only the trivial solution in W̊_{2}^{1}(Ω), we obtain that W σ vanishes in Ω.

Let u^{−} denote the solution of the problem D^{−} with the boundary function σ. We have

V T u^{−} − W u^{−} = u^{−}(∞) in Ω.

Here, by V τ(z) we denote the simple-layer potential with the kernel

Γ(z, q) − Γ(z, 0), q ∈ S\{O}.

Since W u^{−} vanishes in Ω, we have V T u^{−}(z) = u^{−}(∞), z ∈ Ω. The limit relations for the simple-layer potential imply

T (V T u^{−} + \frac{1}{2}u^{−})(p) = 0, p ∈ S\\{O\}.

So one has

V T u^{−}(z) = −u^{−}(z) + u_{0}(z) in Ω^{c},  (32)

where u_{0} is a displacement vector satisfying the boundary condition

T u_{0}(p) = 0 on S\{O}.

From the integral representation of u^{−} and (32) it follows

W u^{−}(z) = u_{0}(z) − u^{−}(∞), z ∈ Ω^{c}.

Since the potential W u^{−} vanishes at infinity, we have

u_{0}(∞) = u^{−}(∞).

The jump formula for W u^{−} implies

σ(p) = u^{−}(p) = u_{0}(p) − u_{0}(∞), p ∈ S\{O}.

Since σ belongs to the class 𝔐, from (32) it follows that u_{0} and its gradient ∇u_{0} grow not faster than a power function as z → 0. For any function ρ ∈ C_{0}^{∞}(R^{2}), satisfying O ∉ supp ρ, we have that ρu_{0} belongs to W_{2}^{1}(Ω^{c}).

From the estimate ∇u_{0}(z) = O(|z|^{−2}) (z → ∞) and Betti’s formula we obtain

∫_{{z∈Ω^{c}, |z|=r}} T u_{0}(z)ds = 0 , 0 < r < δ .

We assume that the Airy function U generated by the displacement vector u_{0} in the domain Ω_{δ} = Ω^{c} ∩ {|z| < δ} satisfies

\frac{∂U}{∂x} + i\frac{∂U}{∂y} = 0 on S_{±}.

In particular, U is constant on S_{+} and S_{−}.

We make the change of the variable t = log(1/r), r = |z|. Let Λ be the image of Ω_{δ} under the mapping (r, θ) → (t, θ), where r, θ are polar coordinates of (x, y). By Γ_{+}, Γ_{−} we denote the images of S_{+} and S_{−}. Then W (t, θ), equal to U(e^{−t}, θ), is a solution of the equation

[(\frac{∂}{∂t} + 2)^{2} + \frac{∂^{2}}{∂θ^{2}} ] [\frac{∂^{2}}{∂t^{2}} + \frac{∂^{2}}{∂θ^{2}} ] W (t, θ) = 0 in Λ

and satisfies W = c_{±}, ∂W/∂n = 0 on Γ_{±}. Moreover, the function W belongs to W_{2,γ}^{2}(Λ) with negative γ. From the local estimate (28) it follows that W belongs to W_{2,γ}^{3}(Λ). Taking into account the relation between u_{0} and U in Ω_{δ} we conclude that W_{0}(t, θ) = (ρu_{0})(e^{−t}, θ)·e^{−iθ} belongs to W_{2,β}^{2}(Λ) with negative β.

Let D(∂_{t}, ∂_{θ}) denote the operator of the boundary value problem

(△^{∗} + K)U = F in Π = {(t, θ) : 0 < θ < 2π, t ∈ R}, T U = f on ∂Π,

continuously mapping W_{2,β}^{2}(Π) into L_{2,β}(Π) × W_{2,β}^{1/2}(∂Π). Here K has the form

K = (\begin{matrix}−λ + 2μ & −(λ + 3μ)(∂/∂θ) \\ (λ + 3μ)(∂/∂θ) & −μ \end{matrix}) .

The eigenvalues of the operator pencil D(ik, ∂_{θ}) : W_{2}^{2}(0, 2π) → L_{2}(0, 2π) × R^{2} are numbers of the form k = iℓ/2, ℓ ∈ Z, and they have multiplicity two. Therefore W_{0} admits the representation

W_{0} = ∑_{k=1}^{p} c_{k}V_{k} + V_{0}

(cf. [8], Theorem 6.2), where p is the largest integer not exceeding 2β, V_{k} are linear independent vector-valued functions satisfying the equation (△^{∗}+K)U = 0 in Λ_{R} = Λ ∩ {t > R} and the boundary condition T U = 0 on ∂Λ ∩ {t > R}. Moreover, we have V_{k} ∉ W_{2,γ}^{2}(Λ_{R}) and V_{0} ∈ W_{2,γ}^{2}(Λ_{R}) for γ ∈ (−1/2, 0). Making the inverse change of the variable r = e^{−t} we obtain

u_{0}(z) = (2μ)^{−1}(κφ_{∗}(z) − z\overline{φ ′_{∗}(z)} − \overline{ψ_{∗}(z)}) + O(|z|^{−ε}),

where ε > −1/2, and

φ_{∗}(z) = ∑_{k=1}^{p} R_{φ}^{(k)}(log \frac{zz_{0}}{z_{0} − z} )(\frac{zz_{0}}{z_{0} − z} )^{(k−p−1)/2} ,

ψ_{∗}(z) = ∑_{k=1}^{p} R_{ψ}^{(k)}(log \frac{zz_{0}}{z_{0} − z} )(\frac{zz_{0}}{z_{0} − z} )^{(k−p−1)/2} .

Here R_{φ}^{(k)} and R_{ψ}^{(k)} are polynomials of degree [(k − 1)/2], and z_{0} is a fixed point in Ω. Since σ ∈ 𝔐, we have u_{0}(z) = O(|z|^{−1/2}). Thus, σ is the restriction to S\{O} of a solution of the homogeneous problem N^{−}.

Theorem 10. Let Ω have an outward peak and let g belong to 𝔑_{ν}, ν > 0. Then equation (1) is solvable in the class 𝔐 and the homogeneous equation has a two dimensional space of solutions. Among solutions of (1) there exists only one with the representation

σ(x) = β_{±}x^{ν−1} + O(1)

if ν ≠ 1/2, and

σ(x) = β_{±}x^{−1/2} log x + O(1)

if ν = 1/2.

Proof. Let V σ be a simple-layer potential with the kernel Γ(z, q). Let u^{+}, u^{−} be solutions of the problems D^{+} and D^{−}. Theorem 1 implies that

T u^{+}(z) = ±β_{0}x^{ν−2} + O(x^{ν−1}), z = x + iy ∈ Ω.

According to Theorem 2, T u^{−} is integrable on S and the displacement vector u^{−}(z) is bounded at infinity. Then

g(p) = V (T u^{+} − T u^{−})(p) + u^{−}(∞), p ∈ S\{O}.

Here the integral is understood in the sense of the principal value.

Let v^{−} denote the solution of the problem N^{−} with the boundary data T u^{+}. According to Theorem 4, we have

v^{−}(z) = O(x^{ν−1}) for ν ≠ 1/2,
v^{−}(z) = O(x^{−1/2} log x) for ν = 1/2, z ∈ S\{O}.

Using the integral representation for w = v^{−} − u^{−} + u^{−}(∞) in Ω^{c} and the jump formula for W w we obtain that

w − 2W w = −2V (T u^{+} − T u^{−}) = −2g + 2u^{−}(∞) on S\{O}.

Thus the displacement vector σ = w −u^{−}(∞) = v^{−} −g is a solution of equation (1).

By Theorem 9 the restrictions of solutions of the homogeneous problem N^{−} to S\{O} (and only such functions) satisfy the homogeneous integral equation of the problem D^{+}. The space of such solutions in 𝔐 is two-dimensional. We choose two linear independent solutions ζ_{R} and ζ_{I} such that

ζ_{R}(z) = ∓\frac{κ + 1}{μ} x^{−1/2} + O(1),

ζ_{I}(z) = ∓\frac{κ + 1}{μ} i x^{−1/2} + O(1), z = x + iy ∈ S\\{O\}.

Therefore solutions of equation (1) in 𝔐 have the form

σ = v^{−} − g − ( Re c)ζ_{R} + (Im c)ζ_{I} .

The complex constant c can be chosen so that σ has the required asymptotics.

4.2. Integral equation for the problem N^{−}.

Theorem 11. Let Ω have an outward peak. Then the homogeneous equation

−2^{−1}τ + T V τ + t_{1}T ϱ_{1} + t_{2}T ϱ_{2} = 0  (33)

has only the trivial solution in the Cartesian product 𝔐 × R^{2}.

Proof. Let τ be a solution of (33). We define the displacement vector v(z) as follows:

v(z) = V τ(z) + t_{1}ϱ_{1}(z) + t_{2}ϱ_{2}(z), z ∈ Ω^{c}.

By integrating (33) over the boundary S we find

∫_{S} τ(p)ds_{p} = 0.

Therefore the potential V τ(z) takes the form

V τ(z) = ∫ {Γ(z, q) − Γ(z, 0)} τ(q)ds_{q} .

Hence we have v(z) = O(|z|^{−1}), ∇v(z) = O(|z|^{−2}), as |z| → ∞. Therefore v(z) is a solution of the problem N^{−} with zero boundary data and

∇v(z) = O(|z|^{β}), β > −1, as z → 0.

The classical uniqueness theorem implies that

v(z) = const, z ∈ Ω^{c}.

Since v(z) vanishes at infinity, it follows that

V τ(z) = −t_{1}ϱ_{1}(z) − t_{2}ϱ_{2}(z), z ∈ Ω^{c}.

We extend the functions ϱ_{k}(z), z ∈ S, k = 1, 2, onto Ω as solutions of the homogeneous Lamé system. Let ϱ_{1}^{+}(z) and ϱ_{2}^{+}(z) denote these extensions. The vector-valued function

ϱ_{0}(z) = V τ(z) + t_{1}ϱ_{1}^{+}(z) + t_{2}ϱ_{2}^{+}(z), z ∈ S\{O}

is bounded in Ω, vanishes on S\{O} and its gradient ∇ϱ_{0} grows slower than a power function. According to Lemma 2, ρ_{0}(z) admits the estimate O( exp(β Re \frac{1}{z})) with negative β as z → 0. We have

V τ(z) = − ∑_{1}^{2} t_{k}ϱ_{k}^{+}(z) + ϱ_{0}(z).

The jump formula for the simple-layer potential entails

τ(z) = − ∑_{1}^{2} t_{k}T ϱ_{k}^{+}(z) + ∑_{1}^{2} t_{k}T ϱ_{k}(z) + T ϱ_{0}(z), z ∈ S.

We have T ϱ_{k}(z) = O(|z|^{−1/2}), k = 0, 1, and T ϱ_{1}^{+}(z), T ϱ_{2}^{+}(z) have order −3/2. Since the components of these functions have different singularity, t_{1} and t_{2} vanish. Thus, V τ(z) = 0 in Ω^{c}. By Lemma 2 applied to V τ in Ω, and by the classical uniqueness theorem we can conclude that V τ(z) = 0, z ∈ Ω. So the density τ given by

τ(z) = T (V τ)^{−}(z) − T (V τ)^{+}(z), z ∈ S\{O},

vanishes on S.

Let ζ_{R} and ζ_{I} be solutions of the homogeneous problem N^{−} introduced in Theorem 10.

Theorem 12. Let Ω have an outward peak, let h belong to 𝔑_{ν}, ν > 0, and satisfy

∫_{S} hds = 0.

Then the integral equation (6) has a unique solution in the direct product 𝔐×R^{2} and this solution can be represented as follows:

σ(z) = β_{±}x^{ν−1} + O(x^{−1/2})  for 0 < ν < 1/2,
σ(z) = (γ_{±} log x + β_{±})x^{−1/2} + O(log x) for ν = 1/2,
σ(z) = γ_{±}x^{−1/2} + β_{±}x^{ν−1} + O(log x)  for 1/2 < ν < 1,

where z = x + iy ∈ S\{O}.

Proof. We consider only one of the statements of the theorem in detail. Let v^{−} denote a solution of the problem N^{−} with the boundary data h such that v^{−}(∞) = 0.

We assume that 0 < ν < 1/2. From Theorem 4 it follows that v^{−} has the form

v^{−}(z) = \frac{1}{2μ} [κφ(z) − z\overline{φ′(z)} − \overline{ψ(z)}] + O(z^{3/2} log z),

where

ψ(z) = δ_{0,0} + δ_{1,0} (\frac{zz_{0}}{z − z_{0}} )^{1/2} + δ_{2,0} \frac{zz_{0}}{z − z_{0}} + β_{0} (\frac{zz_{0}}{z − z_{0}} )^{1+ν} ,

ψ(z) = ε_{0,0} + ε_{1,0} (\frac{zz_{0}}{z − z_{0}} )^{1/2} + ε_{2,0} \frac{zz_{0}}{z − z_{0}} + γ_{0} (\frac{zz_{0}}{z − z_{0}} )^{1+ν} , z_{0} ∈ Ω.

Thus v^{−} admits the representation

v^{e}(z) = c_{0} ± c_{1}x^{1/2} + c_{2}x + d_{±}x^{1+ν} + O(x^{3/2} log x) on S,

where

c_{1} = (1 + κ)δ_{1,0}/2μ, c_{2} = (1 + κ)δ_{2,0}/2μ.

We apply Betti’s formula to v^{−} and first to ζ_{R} and then to ζ_{I} in Ω_{ε} = Ω^{c}\{|z| < ε}. Taking the limit as ε → 0 we obtain

Re c_{1} = \frac{1}{4π} ∫_{S} hζ_{R} ds, Im c_{1} = \frac{1}{4π} ∫_{S} hζ_{I} ds.

We set

t_{1} = \frac{μ}{1 + κ} \frac{1}{2π} ∫_{S} h ζ_{R}ds, t_{2} = \frac{μ}{1 + κ} \frac{1}{2π} ∫_{S} h ζ_{I}ds.

Then the problem N^{−} with the boundary data h − t_{1}T ϱ_{1} − t_{2}T ϱ_{2} has a solution on S_{±} of the form

v(z) = c_{0} + c_{2}x + d_{±}x^{1+ν} + O(x^{3/2} log x).

The solution of the problem D^{+} with the given displacement c_{2}z on S admits the estimate O(z) and this estimate can be differentiated at least once. Hence and by Theorem 1 it follows that the solution u^{+} of the the problem

△u^{+} = 0 in Ω, u^{+} = v^{−} on S,

admits the representation

u^{+}(z) = \frac{1}{2μ} [κφ(z) − z\overline{φ′(z)} − \overline{ψ(z)}] + O(z log z).

differentiable at least once. Here

φ(z) = β_{0}z^{ν} + β_{1}z^{1/2}, ψ(z) = γ_{0}z^{ν} + γ_{1}z^{1/2}.

Therefore the stress T u^{+} has the following representation on S_{±}:

T u^{+}(z) = β_{±}x^{ν−1} + O(x^{−1/2}), z = x + iy.

On S, we have

v = V (T u^{+} − h + t_{1}T ϱ_{1} + t_{2}T ϱ_{2}) .  (34)

The simple-layer potential V T u^{+} is bounded in a neighborhood of the origin and its gradient admits the estimate O(|z|^{ν−1}). Hence and by a classical uniqueness theorem we obtain that (34) remains valid in Ω^{c}. Consequently, the density

τ = T u^{+} − h + t_{1}T ϱ_{1} + t_{2}T ϱ_{2}

satisfies the integral equation of the problem N^{−} and has the required asymptotics. Theorem 11 implies that the constructed solution of equation (6) is unique.

The cases ν = 1/2 and 1/2 < ν < 1 can be studied in a similar way.

Applying Theorems 11 and 12, we obtain the following result.

Theorem 13. Let Ω have an outward peak. Furthermore, let h belong to 𝔑_{ν}, ν > 0, and satisfy

∫_{S} hds, ∫_{S} hζ_{R}ds = 0 and ∫_{S} hζ_{I}ds = 0.

Then equation (2) is uniquely solvable in 𝔐.

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(Received 5.06.2001)

Authors’ addresses:

V. Maz’ya Department of Mathematics Liköping University S-581 83 Linköping Sweden E-mail: vlmaz@mai.liu.se

A. A. Soloviev Department of Mathematics Chelyabinsk State University 454136 Chelyabinsk Russia E-mail: alsol@csu.ru